PT: today for an hour. Seems as if I have knots under the front of my ankle bone and up the hamstring muscles. She did a deep massage. Lets just say the party was over before it even started. ugh Next week if the knots are still there they will do the dry needles into each knot. She says anyway. The guy Chris may not. It's a catch 22 either way. Part 2: Waiting in the lobby was a heavy woman in a wheelchair. The front half of her foot had been amputated and she had metal pins (4 of them) sticking out of both sides of her legs about 6" and 4" above the ankle. She proceeds to tell me a few years ago she had broken a bone in her foot but because of having no feeling due to her diabetic neuropathy she never felt it so she never told her doctor. Well in due time the bone came right out of the skin so she had to have amputation, have pins inserted, move into assisted living and have PT basically forever. Shock: The lady then reaches into her pocket and pulls out two giant Kit Kat bars and proceeds to eat them in less than 5 minutes. I almost passed out in shock! I can still she her cramming those chocolate bars like there was a party and she was the guest of honor. I can say one thing for sure and that is no candy bar, slice of bread, handful of fries, bowl of ice cream or anything else I used to eat on a daily basis will never cross my lips again! Her condition is forever imprinted on my mind. TWO candy bars. Unbelievable!
Sometimes I could just scream and punch somebody!
Why am I so docile? Never taking up for myself. Never saying what I really want to say when somebody insults me or tries to make me feel stupid or uneducated. My feet were red, dry and painful this morning as well as my knees being crushed while I slept....or tried to anyway. Like an a$$ I had to mention it to Sheila who sighed and said "well maybe my legs hurt too and so do a lot of other peoples. " "I said maybe yours hurt from riding a bike down the street after 45 years." Basically she cut me off then and changed the subject and of course I just let sugar drip from my lips as usual like it never hurt my feelings or anything. I think sometimes I have to talk about it just to wrap my head around the fact that this is my new life. Hands are aching but I am typing and later will clean the house even though I will be down then for hours afterward. I need to just embrace the pain as my ankles are already going numb in the front. A few spots on the bottoms of my feet as well and part of the top of my right foot. They say as the numbness continues then the pain is gone. I am thinking I would much rather have the pain than be in a wheelchair in assisted living. perish the thought! On someone who has this could truly understand it. Meds just mask it and make you less aware of your being. Not for me. At least I still have full control of my mentality. I hope they legalize cannabis this year. My doctor said 1 puff 3 times a day would rid my pain for 24 hours. I can't even imagine an entire day without pain. It's been a long time.
Sometimes I get so sick of eating fat. Except fat bombs of course. made them into a nice candy bar yesterday with peanuts. I need some almonds then I could have almond bark. I guess whatever works. I hate this disease. MY leg hurts today from the knee down this morning hopefully can do some stationary bike today if it eases up some. . Hurts? I should say somebody lit a match to it. I believe its the new house slippers which were not cheap by a long shot but I can't hardly wear any kind of shoes so I just keep trying to find some. As for my boots I only need them to get into the car, out and into church, out and back into the car. LOL Keep on keeping on. What else can I do. Exp. continues: Last night ate half a dill pickle, 2 slices of deli roasted turkey breast and 1 oz. of cheese. Still peed all night long but fasting was 120 this morning so I am happy.
"Kooka have you lost your mind?" "No why would you ask that?" "Because hunzies you ordered a pair of knee high boots with heels and I'm afraid now that you are losing your mind." "Well I did put them on and they will look good with my long skirts this winter." "But can you walk in them without your calves going into spasms like they did this morning 2 minutes after you ripped the package open and went prancing around the living room?" "No but I figure if I practice a little each day I may get up to 30 minutes or so." "Kooka. You need medication. You should have had a serious conversation with yourself and come to a more sensible agreement like flat soles maybe." "Just shut up before I punch your lights out!" Last night experiment. Had a snack of pickles slices, small cheese slice and a slice of deli turkey before bed. Sugar 126 this morning so have any carb at all with the snack is not gonna work and besides that I have gained 6 pounds of water in 2 days from that Mexican cheese that I love! Today no cheese.
Ate supper at 7 instead of 5:30. Snack before bed was a slice of deli turkey. Bedtime reading 109 and fasting this morning was 120 so that is a lot better than yesterdays. -15 to be exact. Only woke up once to go to the bathroom so my sleep is so much improved as well as my leg cramps because of the dehydration during the night. I did have to take a prescription Previcid though in the middle of the night as my reflux is giving me a fit right now. Possibly I am drinking was to much coffee or adding so much to my WOE mainly spices. I doubt if it's the ginger because it calms your stomach but the garlic and tumeric could be one of the meanies. Sugars yesterday averaged 110 throughout the day so I am pleased with that and what pleases me most is I have cut my morning Metformin down to a 500 instead of a 1000. Mainly I feel like because I am much ore active during the day and rarely eat any carbs before I eat supper at night. I then take 1000. Been doing that all week and so far it's working. Go me!
Since doing much research I have discovered I have dawn phenonomen. Ex:: Go to be 94, wake up with reading of 120, eat a no carb breakfast and one hour later reading of 102, Two hours after breakfast 98. This continues throughout lunch. Then in the evening it stays steady the rest of the day always going down after lunch and supper but not as fast as after breakfast so my first experiment started last night: 1 2 hours before bed I a 2 ozs. of hard cheese and a small pack of peanuts. Sugar was 110. Woke up at 3:30 a.m. and sugar was 131. Ate an ounce of cheese then woke up at six and sugar was 136. Ate breakfast. One hour later sugar was 115. So eating a no or lock before bed cause the sugar to rise even higher during the night. The only good thing I did not have to pee every 45 mins. as usual when I woke up at 3. My mouth and tongue were not dry as a bone. I did not wake up in a sweat so tonight it's on to exp. #2.
Things are going well today. My foot is an issue. But when is it not. I may have to put the ankle brace back on even though I hate it. I believe I will fry some chicken tonight. Been a while since I made much effort to cook. Cheri told me to smash pork rinds and roll it in them for a nice chrispy coating without carbs. So onward and forward. Sugar was 119 this morning when I got up. God is good. All the time.
last night I gained the world. Oh man did I have a bad night. Finally got a couple hours sleep after I took a Lortab in the middle of the night. I have 2 left and Doc will not renew them even though they expired in 2013. Oh does that make me an addict. I guess so. I said well I guess the gov. wants me to get addicted to pain instead of taking 30 pain pills every 6-9 months of which I only take a half of one since they make me sick to my belly but at least they take the edge off the pain when I can hardly stand it. "Hi there. My name is Barb and I am a pain addict. Nice to meet you. Do you have a good withdrawal program?" Went to a church rummage sale with my aide and we had to leave because my leg started hurting so bad I broke out in a sweat and had to hold onto the wall to make it to the car. Some days are just bum, bum and bummer days. It will be better. Already is because I am sitting and my feet are done cooking for the moment. 2 Cor. 4: 16-18 God is good.
No clue why I am blah. Maybe tired of the humidity. Maybe I want a hot fudge sundae. Maybe I am just bored because I miss Connor. No clue. On a better note riding that portable stationary bike for 30 mins. seems to be helping my leg pain. Sugars staying around 100 during the entire day. Going shopping tomorrow to stock up on some of the things other members are using. My meter is stating a 30 day average at 134. It was 157 for a long minute. Like a year of minutes. I can't complain as I have done this to myself. Not to say I never listened to Doc because I did there were times I just chose to ignore him and my own mind with my I do not give a $#@* attitude. There is always a consequence to every choice.
Diabetic Solutions book by Dr. Bernstein. I can't believe how much good information this book has for type 1 diabetics and type 2. I listed a lot of the foods he suggests and along with the recipes I am off to the grocery again Friday. I did see a few things made from soy flour. I used to make gravy out of flax meal and a bit of heavy cream and water just to have a tad on my eggs in the morning but I am going to go ahead and try it with the soy flour. 3.3 grms. carbs for 1/8 cup and that should make more than enough gravy for me. can't wait. He also suggests those SF 0% calorie syrups. Says he even eats them on his eggs. Yuck! Try the SF jello almost set and then whipped with heavy cream and after it sets drizzle some chocolate SF syrup on it. I am ready to get out of this rut. Sugar down to 100 after breakfast so I am happy.
I cannot believe I waited 3 weeks to go for PT and get these dry needle treatments which as my orthopedic said "will decrease your neuropathy pain especially in your right foot" and the PT says dry needling treatments do nothing for DPN except possibly reduce pain for a few hours. Sooooooooooooo he makes me ride a stationary bike and I am now home with both feet and ankles swollen and a sore knee. I wish I could get the CT on my foot like the x-ray tech said I needed 2 months and 4 doctors ago. I am just so frustrated with the whole thing.
I get so tired of cooking sometimes. I had to force myself to fry an egg this morning. I ate a hot dog on a fork for lunch. Now I have a pot of beef vegetable soup simmering on the stove and will freeze half of it for other days. All the veggys are LC. According to my meter I have my 30 day average down to 136 so onward and forward. Beats prior reading for sure. I can do this. I don't like it but I can and will do it.
Yeehaw I am one happy old chicka. Went to the picnic after church this afternoon. One table loaded with pasta salads, potato casseroles, and the like. Another table baked beans, mac and cheese etc. One table weighted down with cakes, brownies, pies, cookies, apple dumplings and every sweet know to man. A table of hot dogs and hamburgers and one with salad dressings, tossed green salads and melons. So I took my reading 107 and then proceded to eat two burgers no bun, two hot dogs no buns, a spoon of mustard, and a tossed salad with dressing. Two cubes of red watermelon and 2 cubes of yellow watermelon. I am stuffed and my friends are stuffed too.........with sugar. LOL Feeling really pumped about it.
I promised my doctor to work on getting my A1C down in the next four months. If not he is adding another medication to my Metformin. Between the neuropathy pain today and the heat and humidity exercise is the farthest thing from my mind. So is cooking in this weather. I need to plan too cook for the weeks and freeze it and then just nuke it for the hotter days. Oh well maybe I will just eat hot dogs without the bun today. I am bummed.
I used to not care as much as I should have about this ratio given my ignorance of Metabolic Syndrome end points and diabetes, esp. with symptoms I was oblivious to. Keeping an eye on this ratio being below 2.0 as well as other lipid panel results. In general, a bit of a regression over the last month or so which I think showed up in my FBG and most recent A1C numbers. I know I can do better without doing something crazy and unsustainable.
Blood work: Microalbumin/Creatinine ratio elevated at 78 up from 36 two months ago. Wondering if this is due to all of the protein and experiments, or something more sinister like kidney disease and my feeling more tired recently. Will chat with the doctor about this and read up a bit too.
Fasting Blood Glucose was a worry of mine given my recent trends, especially since the last time I "calibrated" against the lab, it was 83 (home) vs 88 (lab). But this time, it was 87 at the lab and 96 at home. <shrug> I have a lot of silly graphs here.
A1C is at 6.0 - certainly an improvement vs. the 13.2 at DX 3 months ago, but I'd like to go a lot lower and more importantly, figure out what's sustainable.
Some thoughts on this ~90th day after DX: FBG still okay relative to that 198 DX FBG value, which was insane and panic'd me quite a bit, along with the 13.2 A1C, but need to keep an eye on this upwards trend and upwards trends in general - Honeymoon Period, "The Easy Part", or similar effects, T1/T1.5/LADA actual DX, increasing my carb intake far too fast, too soon, and some of the other things which correlated with the start of at least one trend - being sick, that crazy fitbit challenge I did for 2 weeks, backed off on the hard boiled eggs (no good reason, I still LOVE them), stopped kale consumption, greatly reduced veggie consumption, HUGE increase in cashew consumption, no more "super" low carb days.
The idea of fasting (intermittent, or whatever the terminology, where things are not drastic, yet still quite deliberate) - intrigues me more and more despite being the opposite of what some have written. I think fasting also fits my personality quite well. Will have to add that to the toolbox.
Speaking of The Toolbox, haven't had to use the monster muscle groups yet and related routines (proper squats, deadlifts; compound movements, and so on). As needed. Later.
Do want to think more about portion control. I feel like a pig sometimes and it just doesn't seem okay anymore. Believe this is a reasonable, sustainable change and I've noticed I actually take home doggie bags far more often. That never happened in the past.
Totally forgot to keep track of something potentially more meaningful than weight, like my waist size, which has gone down enough for me to buy new jeans and forced me to buy a new belt since I was beyond the last hole.
Still continuing to really ignore flexibility and stretching. Will have to work this in. Really bad things have happened before when this has been ignored.
4 small eggs fried in ghee and red palm oil* spinach kimchi *no orangutans were harmed. Nutiva brand sourced in Ecuador. ------------------------------------------------------- Only did a 36 hr fast this time. May need more time between day+ fasts. I may have to try IF again, although in past I've found it harder than longer fasts. hmmm Experimentation needed. Goal - max weight and BG level reduction along with mood raising effects, versus discomfort level and time. ------------------------------------------------------ Managed to eat slightly more than half that plate. Eggs must be the most filling food ever. Lunch leftovers along with a bit from last nights dinner.
Had a lovely lunch with my brother, ate 1/2 my plate of shaking beef (saved rest for dinner (that didn't happen)). Went back home, helped him haul about various furniture and stuff he was picking up and wound up taking my BG at 3 hrs PP. 133! Damn it. Apparently sugar in the marinade. A plate full of beef, watercress and a few tomato slices shouldn't do that. Unless it was the effect of me staring mournfully at his Vietnamese coffee full of condensed milk. (not that mournfully, I KNEW what that coffee would do) I can't think of anything else at that restaurant that might be safer. This could be a problem since the whole rest of the family LOVES the place. I'll have to check if they can customize something for me. When I test for ketosis at the end of my fast, the Ketonix (breathalyzer) said I hadn't made it. (after a full 4 days!) Either it's not accurate or my liver is a really busy little devil. I'm debating doing a fat fast and then back into a few days of fasting to try to get into keto. On the other hand, the last 2 days of my fast I felt really great and energetic (except for the short bit where I forgot to drink the salty broth.) That's usually a pretty good sign of ketosis. I might have to spring for the keto blood test strips. On a happy note, AC is coming by Friday. Major duct cleaning today and installation the next 2 days. HALLELUJAH!
I decided to break my fast last night when I remembered my brother is due to come by today. And that means we will go to his favorite Vietnamese restaurant for Shaking Beef. Yum. Had a very tiny dinner last night - 1/3 cup greek yogurt (2% ugh) but it was in the fridge and yogurt recommended as a first food for breaking fast. Had a small reaction to the yogurt TMI ALERT - small watery bowel. Apparently typical but unpleasant when breaking multi day fast but ewwwww. Will be prepared next time Waited 1/2 hour and ate 2 oz of trout and 1/2 cup greek salad. 5.5 gms carbs. Tiny meal but I was very full. Overall I'd say the fast experience was good and I'll do it again. My energy was excellent on the third day. Yesterday energy was down in the AM and I took a nap. After that I felt great and would have continued, but brother visiting and shaking beef is too good to pass up. 7/27 12pm 111, 6pm before dinner 95, 2 hrs PP 113 (gotta love my liver!), started 500 mg met in evening, before bed 103 7/28 FBG 104 2 eggs, 1/2 avocado and salsa verde, black coffee this morning 8.5 carbs, 3.4 net. Getting used to eating again slowly. I don't know how fast metformin works but between it and the fast and lc (and not binging on ice cream!) my fbg is down 102 points since last Monday.
I haven't been here for a while and frankly fell hard off the low carb wagon during that time. Depression, anxiety and life hit me hard for a while. Monday a week ago I got a serious wakeup call in the form of a FBG of 206 and an A1C of 8.3. I'd also shot up to 202 pounds from the low of 170 I'd gotten to on LC a year and a half ago. Doc put me on Metformin HCL 500 one in AM and starting today one in AM and one in PM. (lucky me, none of the digestive Met problems) I went straight back to eating a LC diet and walking and weights. I'm kind of panicking though that I let things get this bad. I'm very lucky that so far I don't seem to have any of the horrid D side effects, although of course, bad things are bound to be happening inside my body. So off to digging through the depths of the internet, hoping to sort the gold from the dross. So I came across this site (URL deleted because I think this got me in trouble before) with heaps of info from a Dr. Jason Fung whose emphasis is on reducing the insulin resistance that is the hallmark of T2 diabetes. One of the methods he emphasizes to get that down is fasting, both intermittent and longer fasts. Sounds like a plan after reading through how he recommends doing fasting. His site is fascinating - lots of videos on diabetes, obesity, insulin resistance, fasting etc. Articles also. I'm starting day 4 now. Non-caloric beverages - water, tea, coffee, no sweeteners or sodas. Bone broth. (I add salt, a bit of potassium and magnesium to the broth) A squirt of lemon juice ok, a Tbs of chia seeds once a day ok, a Tbs of coconut oil ok. It's not been bad and my BG has been dropping fast, down to a low of 100 at bedtime night before last. Dawn phenomenon hit hard this am though and I had a 129. Ugh. I'm hoping to continue this until I can get down to at least one reading of 83, then go back to VLC (Bernstein style) and IF or periodically do longer fasts. Otherwise I'll stop when either serious hunger hits again or anything like lightheadedness or lower BP. So far I'm feeling pretty fantastic - clearer head, more energy, more motivation and the depression/anxiety is way down. 7/20: FBG 206! Started Metformin Started eating LC again 7/21 Eating LC No BG, couldn't find meter 7/22 FBG 158, Breakfast - 2 hrs PP 189 (hands may have been dirty - out shopping), 2.5 hrs PP 159, Lunch, 2 hrs PP 152, Dinner, 2 hrs PP149 7/23 FBG 154, 2 hrs later - (only coffee) 176 holy #@$%, breakfast, 2hrs PP 124. lunch, 3 hrs PP 131, dinner 2 hrs PP 147 7/24 Start of Fast - FBG 145, 12pm 134, 6pm 115 7/25 6am 130, 12pm 131, 6pm 111, 9pm 100 7/26 6am 112, 12pm 110, 6 pm 104, 10pm 111 7/27 6am 129 I'll try to post daily to stay accountable and to have a handy place to keep notes. Feel free to comment if you want.
Seems like this blog area is the right place for this sort of entry vs. my first post on this DF site. Yikes! (though all of you were quite accommodating and just went with the flow) Anyhow, I did want to snapshot a entry after my first 30 days after DX and see how things change over time. I've already uploaded my FBG chart in my images gallery but here are some raw thoughts at Day 30. What seems to help, at least as I look forward from today.... Ultimately focusing on doing whatever it takes to be strong, supportive and to be there for my family.
Considering the hardships so many others face each and every day which make my situation look like a walk in the park. Think about mom's cancer situation and how far she went with so little education, so little support, horrible doctors, and so on.
My love of numbers, measuring things, hitting targets. It's a large part of we as engineers do to solve problems. Fitbit, blood works, fitness metrics, various vitals, glucose meter and carb counting - all align very nicely with this.
The act of writing. Putting thoughts down on (virtual) paper, via forums, etc. - and the sense of community via WWW online forums.
What I worry about.... My family worrying about me. Stress on my family.
Fatigue and low energy due to diet changes.
Getting sick (or throwing out my back, etc.) for a long enough period where things can't be controlled as much via exercise. Perhaps diet will have to be enough. Or diet plus meds.
What seems to be working for me, or at least what resonates with me.... There has definitely been a huge reduction in carbs and also almost complete elimination of junk and processed foods. That said, I like the idea of not eating as many carbs if I know I cannot work out. Eating a bit more if I feel like it and I know I will work out afterwards. Exercise cannot always overcome diet, at least given what I'm not willing to do on the exercise front.
Not eating too close to bedtime. Letting my body enjoy the low BG as I sleep, even though I may encounter some DP situations. In the same vein, take a nap only after measuring a low BG level. This seems to have been demonstrated in my real-world data points.
Strength training is important in terms of the pump and for building muscle.
Don't let perfection become the enemy of the good enough. Can't always achieve a perfect schedule or pure sustainability. Life gets in the way. shoot happens. Need to adapt, use your creativity, intellect, and figure things out. Take what you can get, when you can get it (vs. only looking at 100% sustainable solutions) - this is especially true for exercise.
Not eating if I'm not hungry (i.e. social situations, "time for lunch" routines, etc.).
Okay to do short fasts. Related to the bedtime rule, but so far, I haven't done anything beyond that. Adding this here as a contrast to my prior eat-every-2-hours brainwashing.
Proper management of sleep, stress, exercise, and diet seem like the way to go - the usual suspects!
[earlier] had some mild but clear symptoms for a few months, did some exercise, symptoms went away; but was worried enough to see a doctor given what I had read online about diabetes; diet up until this point was LOADED with carbs (200+ grams of carbs per sitting quite common) and close to 100% restaurant, fast food, and all-you-can-eat catered food and snacks. DX'd with Metabolic Syndrome ages ago, no symptoms, didn't pay attention, bad move. 2015.05.28 DX'd A1C at 13.2, FBG at 194 - The Wake Up Call !! 2015.06.08 [11 days after DX ] FBG mostly in 80s, all below 100; routine and details below; visual graph in my gallery 2015.06.27 [~1 month after DX] lab A1C at 9.8 & FBG at 88 ; lipid and metabolic panels 100% normal 2015.07.28 [~2 months after DX] lab A1C at x.x 2015.08.28 [~3 months after DX] lab A1C at x.x Diet: low-to-moderate carbs; eat a ton; love my fats, especially given my love of hard boiled eggs, all kinds of cheeses and meat; though I get a lot of veggies in too; much more mindful of overeating and also try to only eat when hungry Exercise: daily mild, but deliberate walking (~12,000 steps a day on average, very light strength training, occasional but well beyond recreational tennis)- intimately familiar with HIIT and various conditioning routines, but not going down that path for now; that said, I do need to spend WAY more time on stretch/flexibility routines - otherwise really bad things happen Medication: currently not taking any medication for diabetes or non-diabetes Weight: as of 2015.06.28 roughly at 200 pounds, down from an all time high of ~250 not that long ago this is technically the lightest I've been in since at least March 7, 2005 (the earliest records I have) - as a 6' male, given my activity level, athletic exposure, and genetics, need to figure out a reasonable goal given my muscle mass, standards, sustainability, and age Way too early to talk about sustainability, but so far, things have been encouraging, with minimal signs of craving, fatigue, cheating, etc. Though being so new, all of this requires much more homework, understanding, measuring, experiments, and so on. Also rough for me to be away from ridiculous amounts of rice, pasta, ramen noodles, and so on. I grew up with that stuff!
Trying to get Pregnant with a Broken Down Body!


Manually updating menus throughout a store or multiple locations can be time consuming. Chobani SoHo needed a POS system that could quickly update menu changes across all connected devices, including mobile order takers.
Being in the quick service industry, speed of service is crucial to Chobani SoHo's success. Revel's iPad POS keeps up with high volume demand, never hitting a hitch when it comes to processing sales during busy times.
Chobani is America's number one Greek yogurt. Since the company started in 2005, founder and CEO, Hamdi Ulukaya, knew he wanted to open a physical location to show off the brand, yogurt, and deliver a healthier lifestyle. In 2012, after booming success, Chobani opened its flagship cafe in New York City's popular SoHo neighborhood and titled it Chobani SoHo with Revel iPad Point of Sale solution. In addition to serving their famous Greek yogurt, Chobani SoHo also serves hot drinks and fresh sandwiches inspired from the Mediterranean. Chobani in NYC uses iPad POS so should you.
"On our busiest summer day, we'll have around 600 to 700 people come through our doors. Average wait times tend to be about 2 to 3 minutes. Revel's been really great on pushing those numbers down and helping us move lines because we can utilize efficient hardware, such as Kitchen Display Systems (KDS), and features like mobile ordering. It's really awesome."
Chobani chose Revel for it's all-in-one solution, with flexible software and top-of-the-line hardware to maximize efficiency. By using Revel's Mobile Order Takers and Kitchen Display System alongside their iPad POS terminals, Chobani has streamlined operations, increasing the speed of order processing and serving customers as fast as possible. Chobani looks forward to growing internationally with Revel by its side.


The Gallery at Queen's Park presents a selection of works from the Plaskett Collection on loan from the New Westminster Public Library. This exhibition is in connection with the Arts Council of New Westminster's 50th Anniversary Celebration during the month of November, 2017.
Also included in the celebration will be other events, including Analog 3.0 held at the River Market, and a secret concert at The Gallery at Queen's Park on November 29th. Join us for The ACNW 50th Anniversary Celebration Reception at the Centennial Lodge held on November 26th, from 2:30 – 5:00pm.
Joseph "Joe" Plaskett (12 July 1918 – 21 September 2014) was a Canadian painter. He studied with many prominent Canadian painters like A.Y. Jackson, Jack Shadbolt, Lawren Harris and Jock Macdonald. Plaskett was a pupil of Hans Hofmann in New York and Provincetown in 1947 and 1948.
Plaskett was born in New Westminster, British Columbia in 1918. In 1950, he arrived in Paris where he studied with Fernand Léger, and Jean Lombard, etching and engraving with Stanley William Hayter. He taught intermittently in Canada until 1957. He then settled in Paris where his studio became an informal salon for Canadian painters, writers, poets and filmmakers, interfacing with artists from other countries.
Plaskett studied art in Banff, San Francisco, New York, London and Paris. He lived in Paris from 1951 until shortly before his death when he moved to Suffolk, England. His chosen subjects were always intimate expressions of everyday life – interiors, still life, and portraits of friends and models.
Plaskett continued to paint daily until his death on 21 September 2014 at the age of 96.


text:
\section{Introduction}

In the last decade, Leavitt path algebras have been generating significant interest. Introduced in \cite{Gene_Gonzalo1} and \cite{AMP}, these algebras represent an algebraic analog of a class of $C^*$-algebras. Leavitt path algebras are free algebras over a field that satisfy the same relations as the graph $C^*$-algebras (the introduction to M. Tomforde's paper \cite{Tomforde} is a good source for more details on graph $C^*$-algebras). In addition to being algebraic counterparts of graph $C^*$-algebras, Leavitt path algebras are generalization of the classical Leavitt algebras, the algebras that fail the invariant basis number property universally in a certain respect (for more details see \cite{Bergman}).

In \cite{AB}, a row-finite Leavitt path algebra $L(E)$ is embedded in a (von Neumann) regular algebra $Q(E)$ with isomorphic monoids of isomorphism classes of finitely generated projective modules. The algebra $Q(E)$ is called the regular algebra of a Leavitt path algebra. In this paper, we describe exactly when $Q(E)$ is unit-regular, self-injective, directly finite, and equal to the maximal (left and right) ring of quotients of $L(E)$. We relate these conditions with those stating that a Leavitt path algebra is right and left noetherian, directly finite and without infinite set of orthogonal idempotents.

By \cite[Theorems 3.8 and 3.10]{AAS2}), it is known that the Leavitt path algebra $L(E)$ over a graph $E$ is (left and right) noetherian exactly when $E$ is finite and no cycle in $E$ has an exit. This last condition is known as (NE) in the literature. In this paper the graphs satisfying Condition (NE) are called no-exit graphs for short. We obtain further characterizations of noetherian Leavitt path algebras as well as the characterizations of the above mentioned algebraic properties of $Q(E)$ over a finite no-exit graph $E$. In particular, we obtain a set of nine new equivalent conditions relating many algebraic properties of $L(E)$ and $Q(E)$ to the no-exit condition on a finite graph $E$ (Theorem \ref{equivalences1-14}). Moreover, if the involution on the base field $K$ is positive definite, we also prove that the involution of $L(E)$ extends to $Q(E)$ making $Q(E)$ $^\ast$-regular, symmetric and finite just in case that $E$ is a finite no-exit graph (Theorem \ref{equivalences15-19}).

After that, we consider when $Q(E)$ is equal to the maximal, total and classical left, right and symmetric ring of quotients of $L(E)$ adding another ten equivalent conditions (Corollary \ref{Q(E)_as_quotient} and Proposition \ref{extendible_LPAs}) to the list of the equivalences.
A Leavitt path algebra of a finite no-exit graph is hereditary and noetherian so it may not be surprising that it has all the properties mentioned above. However, we emphasize the fact that from our results it follows that a Leavitt path algebra enjoys all these properties {\em only} if the underlying graph is finite and without exits.

Using some of these results, we obtain an explicit description of the inverse of the isomorphism $V(L(E))\rightarrow V(Q(E))$ of monoids of equivalence classes of finitely generated projective modules over a Leavitt path algebra and its regular algebra (Theorem \ref{K0_theorem}).

Finally, we give some generalizations to the isomorphism conjectures posed in \cite[Conjecture 1]{Abrams_Tomforde}, and show that, for the class of Leavitt path algebras considered in this paper, a strongly generalized version of the Isomorphism Conjecture of Graph Algebras \cite[p. 22]{Abrams_Tomforde} holds. Concretely, for the family of noetherian Leavitt path algebras, if $L(E)$ and $L(F)$ are isomorphic as rings, then they are isomorphic as $^\ast$-algebras as well. This, in turn, expands the family of graphs for which a positive answer has been given to the Isomorphism Conjecture for Graph Algebras. Specifically, if $E$ and $F$ are finite no-exit graphs, then $L_{\mathbb C}(E)\cong L_{\mathbb C}(F)$ as rings implies that $C^*(E)\cong C^*(F)$ as $^\ast$-algebras.

The paper is organized as follows. In Section \ref{section_recap}, we recall the main definitions, the construction of the regular algebra and some existing results. In Section \ref{section_charac_noeth}, we prove the equivalence of fourteen  conditions describing algebraic properties of $L(E)$ for a finite no-exit graph $E$ and of its regular algebra $Q(E)$. In Section \ref{section_symmetric}, we describe $Q(E)$ as a ring of quotients of $L(E)$ and add additional three equivalences to previously obtained fourteen. In Section  \ref{section_K_0}, we add seven final conditions to the set of equivalences and use them to describe the inverse of the map $V(L(E))\rightarrow V(Q(E))$ of monoids of equivalence classes of finitely generated projective modules. Lastly, in Section \ref{section_isomorphism_conjecture} we answer in the affirmative both the Isomorphism Conjecture for Graph Algebras and the Strongly Generalized Isomorphism Conjecture for the class of noetherian Leavitt path algebras.

\section{Summary of the construction of the regular algebra and some related results}
\label{section_recap}

Throughout this paper, $K$ denotes a field and $E$ a directed graph. We write $E=(E^0, E^1, r, s)$ to denote that $E^0$ is the set of vertices, $E^1$ the set of edges, and $r$ and $s$ maps $E^1\rightarrow E^0$ describing ranges and sources of edges respectively. We say that $E$ is finite if both $E^0$ and $E^1$ are finite and that $E$ is row-finite if every vertex emits only finitely many edges ($|s^{-1}(v)|<\infty$ for all $v\in E^0$). A vertex $v$ is called a sink if $s^{-1}(v)=\emptyset.$ It is called a source if $r^{-1}(v)=\emptyset.$

A path of length $n$ is a sequence of edges of the form $p=e_1\ldots e_n$ for some positive integer $n$ such that $s(e_{i+1})=r(e_i)$ for $i=1,\ldots, n-1.$ In this case we define the source $s(p)$ of $p$ to be $s(e_1),$ and the range $r(p)$ of $p$ to be $r(e_n).$ If $s(p)=r(p),$ $p$ is said to be closed. If $p$ is closed and $s(e_i)\neq s(e_j)$ for $i\neq j,$ then $p$ is called a cycle. An edge $e$ is an exit of a path $p=e_1\ldots e_n$ if there exists $i$ such that $s(e) = s(e_i)$ and $e\neq e_i.$

Considering vertices to be the paths of length 0, let $E^*$ denote the set of paths of all non-negative lengths. The path algebra $P_K(E)$ is a free $K$-algebra over the set of all paths $E^*$ where the multiplication of paths $p$ and $q$ is the concatenation if $r(p)=s(q)$ and it is 0 otherwise. Alternatively, $P_K(E)$ is a free $K$-algebra with basis consisting of vertices and edges of $E$ such that
\begin{itemize}
\item[(P1)] $vv=v$ and $v w=0$ if $v\neq w,$
\item[(P2)]  $e=s(e)e=er(e),$
\end{itemize}
for all vertices $v$ and $w$ and all edges $e.$ The notation $P_K(E)$ is shortened to $P(E)$ when we work over the same (fixed) field.

For a given graph $E,$ consider an extended graph of $E$ to be the graph with the same vertices and with edges $\{e\ |\ e\in E^1\}\cup \{e^*\ |\ e\in E^1\}$ where the range and source relations are the same as in the original graph for $e\in E^1$, and $r(e^*)=s(e)$ and $s(e^*)=r(e)$ for the added edges. The edges $e^*$ are called ghost edges. The Leavitt path algebra $L_K(E)$ is the free $K$-algebra with basis consisting of vertices, edges and ghost edges that satisfies the path algebra axioms (P1) and (P2) in addition to
\begin{itemize}
\item[(CK1)] $e^*e=r(e),$ and $e^*f=0$ if $e\neq f$ for all $e,f\in E^1,$

\item[(CK2)] $v=\sum ee^*$ for all $e\in E^1$ with $v=s(e)$ and all $v\in E^0$ with $0<|s^{-1}(v)|<\infty.$
\end{itemize}
These last two axioms are called  Cuntz-Krieger relations. The notation $L_K(E)$ is also often shortened to $L(E)$ when there is no danger of confusion.

It a well-known fact that $L(E)$ is unital with the identity element $1=\sum_{v\in E^0} v$ if and only if $E^0$ is finite (e.g. see \cite[Lemma 1.6]{Gene_Gonzalo1}) and that $L(E)$ is a $\mathbb{Z}$-graded $K$-algebra, spanned as a $K$-vector space by $\{pq^{\ast } \ \vert \ p,q \text{ are paths in }E\}$. (Recall that the elements of $E^{0} $ are viewed as paths of length $0$, so that this set includes elements of the form $v$ with $v\in E^{0}$.) In particular, for each $n\in\mathbb{Z}$, the degree $n$ component $L_{K}(E)_{n}$ is spanned by $\{pq^{\ast }\ \vert \ p,q \text{ are paths in }E\text{ with } l(p)-l(q)=n\}$, where $l(p)$ denotes the length of $p$.

Note that $L(E)$ is a ring with involution. Namely, for an involution $k\mapsto \overline{k}$ of the field $K$ (which may be taken to be the identity), one can define $(\sum k_{p,q} pq^*)^*=\sum \overline{k_{p,q}}qp^*.$ In \cite[Lemma 1.3.1]{Work}, I. Raeburn notes that, for the class of row-finite graphs without sources, if the involution on $K$ is {\em positive definite} (i.e. $\sum_{i=1}^n \overline{k_i}k_i=0$ implies $k_i=0$ for all $i=1,\ldots,n$ and for all non-negative integers $n$), then the involution on $L(E)$ is {\em proper} (i.e. $x^{\ast}x=0$ implies $x=0$). In \cite[Proposition 2.3]{ARV} it is shown that this result holds for all graphs.

The algebra $L(E)$ is always nonsingular as a ring (\cite[Proposition 4.1]{Mercedes_quotients}) and it is hereditary if $E$ is finite (\cite[Theorem 3.5]{AMP}). In the last decade, numerous ring theoretic properties of $L(E)$ (for example being finite-dimensional, simple, semisimple, purely infinite simple, regular, noetherian, artinian, exchange, prime, primitive, to name just a few) have been characterized in terms of graph theoretic properties of the underlying graph $E.$ We recall one such result.

Recall that a graph $E$ is said to satisfy Condition (NE) (NE for ``no-exit''), or that $E$ is a \emph{no-exit graph} for short, if no cycle in $E$ has an exit.
Also, recall that a ${\mathbb Z}$-graded $K$-algebra $A=\bigoplus_{n\in \mathbb{Z}} A_n$ is \emph{locally finite} in case $\dim_K(A_n)<\infty$ for every $n\in {\mathbb Z}$.

\begin{theorem} {\bf (\cite[Theorems 3.8 and 3.10]{AAS2})} For a finite graph $E$ and field $K$ the following conditions are equivalent:
\begin{enumerate}
\item $E$ is a no-exit graph.
\item $L(E)$ is left noetherian.
\item $L(E)$ is right noetherian.
\item $L(E)$ is locally finite.
\item If $l$ is the number of cycles in $E$ (call
them $c_1,\dots,c_l$), $m_i$ the number of paths ending in a fixed (although arbitrary) vertex $v_{m_i}$ of
the cycle $c_i$ which do not contain the cycle itself (for $1\leq i \leq l$), $k$ is the number of
sinks in $E$ (call them $w_{l+1}, \dots, w_{l+k}$), and for every $j\in \{1, \dots, k\}$, $n_j$
is the number of paths ending in the sink $w_{l+j}$, then $L(E)$ is isomorphic to a direct sum of the matrix rings below
$$\left(\bigoplus_{i=1}^l M_{m_i}(K[x, x^{-1}])\right)\oplus
\left(\bigoplus_{j=1}^{k}M_{n_j}(K)\right).$$
\end{enumerate}
\label{noetherian_NE_theorem}
\end{theorem}
This result was generalized for locally noetherian Leavitt path algebras over row-finite graphs in \cite[Theorem 3.7]{NE-paper}.

Lastly, we recall the concept of the regular ring of a (Leavitt) path algebra. In \cite{AB}, the regular ring $Q_K(E)$ of a path algebra $P(E)$ (and its Leavitt path algebra $L(E)$) is constructed. We outline the main idea of the construction from \cite{AB}. In the following, we fix the field $K$ and shorten the notation $Q_K(E)$ to $Q(E).$

\begin{enumerate}
\item[(i)] Let $E$ be a finite graph. Let $\Sigma$ be the set of matrices with entries in the path algebra $P(E)$ that become invertible in the algebra of power series $P((E))=\{\, \sum k_pp\, |\, p\in E^*$ possibly infinitely many $k_p$ are nonzero $\}.$

The universal localization of $P(E)$ with respect to $\Sigma$ (obtained by adding the entries of the inverse matrices of matrices in $\Sigma$ to $P(E)$) is the division and the rational closure of $P(E)$ in $P((E))$ (see \cite[Observation 1.18 and Theorem 1.20]{AB}). In \cite{AB}, it is denoted by $P_{rat}(E).$ By \cite[Proposition 2.15]{AB}, $P_{rat}(E)$ is always semihereditary.

\item[(ii)] Let $\Sigma_1$ be the set of the following homomorphisms of finitely generated projective left $P(E)$-modules. For every non-sink vertex $v,$ let $e_1,\ldots e_n$ be all the edges that $v$ emits. The homomorphism $\mu$ mapping $P(E)v$ to $P(E)r(e_1)\oplus\ldots\oplus P(E)r(e_n)$ by $r\mapsto (re_1,\ldots, re_n)$ is in $\Sigma_1$ and the algebra $L(E)$ is the universal localization of $P(E)$ with respect to $\Sigma_1.$

Also, let $\overline{E}$ be the opposite graph of $E,$ i.e. the graph with $\overline{E}^0=E^0,$ $\overline{E}^1=\{\overline{e}\ |\ e\in E^1\},$ $\overline{s}(\overline{e})=r(e)$ and  $\overline{r}(\overline{e})=s(e).$ Let $\Sigma_2$ denote the set of the following homomorphisms of finitely generated projective left $P(\overline{E})$-modules: for every non-sink vertex $v,$ let $e_1,\ldots e_n$ be all the edges that $v$ emits and $\nu$ be the mapping $P(\overline{E})r(e_1)\oplus\ldots\oplus P(\overline{E})r(e_n)$ to $P(\overline{E})v$ by $(r_1,\ldots, r_n)\mapsto \sum r_i \overline{e}_i.$ Then $L(E)$ is the universal localization of $P(\overline{E})$ with respect to $\Sigma_2.$

\item[(iii)] The ring $Q(E)$ is the universal localization of $P_{rat}(E)$ with respect to $\Sigma_1.$ It is also the universal localization of $L(E)$ with respect to $\Sigma.$ Finally, it is also universal localization of $P(E)$ with respect to $\Sigma\cup\Sigma_1.$

Moreover, if $E$ has $d$ vertices, the following diagram commutes.
$$\xymatrix{
K^d\ar[d]\ar[r]&P(E)\ar[r]^{\Sigma^{-1}}\ar[d]^{\Sigma_1^{-1}}&P_{rat}(E)\ar[r]\ar[d]^{\Sigma_1^{-1}}&P((E))\ar[d]^{\Sigma_1^{-1}}\\
P(\overline{E})\ar[r]^{\Sigma_2^{-1}} & L(E)\ar[r]^{\Sigma^{-1}}  &  Q(E)\ar[r] & U(E) }$$
Here the algebra $U(E)$ is the universal localization of $P((E))$ with respect to $\Sigma_1^{-1}.$

\item[(iv)] The ring $Q(E)$ satisfies the path algebra axioms (P1), (P2), together with (CK1) and (CK2).

\item[(v)] The ring $Q(E)$ is regular, hereditary and such that the monoids of finitely generated projectives $V(L(E))$ and $V(Q(E))$ are isomorphic (see \cite[Theorem 3.5]{AMP} and \cite[Theorem 4.2]{AB}).

\item[(vi)] The algebra $L(E)$ is a perfect right ring of quotients of $P(E)$ and a perfect left ring of quotients of $P(\overline{E})$. The ring $Q(E)$ is the total left ring of quotients of $P(\overline{E})$ and total left ring of quotients of $L(E)$ (all the necessary background on rings of quotients can be found in \cite{Stenstrom}). So we have that
\[Q(E)=\Qltot(L(E))=\Qltot(P(\overline{E}))\]
The proofs of these claims can be found in \cite{Miquel}.

\item[(vii)] If $E$ is row-finite (but not finite necessarily), $Q(E)$ is the direct limit of the regular rings of path algebras of finite subgraphs of $E$ (\cite[paragraph before Theorem 4.4]{AB}).
\end{enumerate}

\section{Characterizations of noetherian Leavitt path algebras}
\label{section_charac_noeth}

In this section, $E$ denotes a finite graph. First, we prove two sufficient conditions for $E$ to be no-exit. Then, we prove some necessary conditions for $E$ to be no-exit. As a corollary, we obtain a list of equivalences to the condition that $E$ is no-exit. After that, we study some implications of the statement that the involution extends from a Leavitt path algebra to its regular algebra. As a corollary, we obtain further characterizations of noetherian Leavitt path algebras in case when the involution on underlying field is positive definite.

\begin{proposition} Any of the following two conditions imply that $E$ is a no-exit graph.
\begin{itemize}
\item[(i)] $L(E)$ is finite.

\item[(ii)] $L(E)$ contains no infinite set of orthogonal idempotents.
\end{itemize}
\label{sufficient}
\end{proposition}
\begin{proof} (i) Let us assume that $L(E)$ is finite but that $E$ has a cycle $p$ with an exit $e$. By rotating the cycle if necessary, we can assume that the exit $e$ occurs at the base of the cycle $p$, so that $p=e_1e_2\dots e_n$ and $v=s(e_1)=s(e)$ with $e_1\neq e.$ Consider the element $x=p+\sum_{w\neq v}w$ that satisfies: $x^*x=(p^*+\sum_{w\neq v}w)(p+\sum_{w\neq
v}w)=p^*p+\sum_{w\neq v}w=v+\sum_{w\neq v}w=1.$ Then $xx^*=(p+\sum_{w\neq v}w)(p^*+\sum_{w\neq
v}w)=pp^*+\sum_{w\neq v}w=pp^*+1-v=1$ and so $pp^*=v.$ Multiplying by $e^*$ on the left we
have $0=(e^*e_1)(e_2\dots e_n)p^*=e^*pp^*=e^*v=e^*\neq 0$, a contradiction. Thus, there cannot be a cycle with an exit and so $E$ is a no-exit graph.

(ii) Suppose, by way of contradiction, that $E$ is not a no-exit graph so that there exists a
cycle $p$ with an exit $e$. By relabeling the vertices if necessary, we can assume that $s(e)$ is the base of
the cycle so that we can write $p=e_1\dots e_n$ with $s(p)=s(e_1)=s(e)$ and $e_1\neq e$. In this case we
clearly have $e_1^*e=0=e^*e_1$ which in turn implies $p^*e=e^*p=0$. Consider the set $\f=\{p^nee^*(p^*)^n \}_{n=1}^{\infty}.$ It is easy to check that the elements of $\f$ are orthogonal idempotents. However, $\f$ is infinite. To see that let $n>m$ and assume that $p^nee^*(p^*)^n=p^mee^*(p^*)^m.$ Multiplying by $e^*(p^*)^m$ on the left, we obtain
$$0=(e^*p^{n-m})ee^*(p^*)^n=e^*(p^*)^mp^nee^*(p^*)^n=e^*(p^*)^mp^mee^*(p^*)^m=e^*(p^*)^m,$$ a contradiction
with the fact that ghost paths are linearly independent in $L(E)$ by \cite[Lemma 1.6]{G2}. Thus, we obtain an infinite set of orthogonal idempotents, a contradiction with the hypothesis.
\end{proof}

\begin{proposition} The condition that $E$ is a no-exit graph implies any of the following two conditions.
\begin{itemize}
\item[(i)] $Q(E)$ is unit-regular, (left and right) self-injective and $Q(E)=\Qltot(L(E))=\Qlmax(L(E)).$

\item[(ii)] If $\phi$ is the isomorphism
$$\phi: L(E)\cong \left(\bigoplus_{i=1}^l M_{m_i}(K[x, x^{-1}])\right)\oplus
\left(\bigoplus_{j=1}^{k}M_{n_j}(K)\right)$$ as in Theorem \ref{noetherian_NE_theorem}
where $l$ is the number of cycles  $c_1,\dots,c_l$  in $E$, $m_i$ the number of paths ending in a fixed vertex $v_{m_i}$ of
the cycle $c_i$ which do not contain the cycle itself for $1\leq i \leq l$, $k$ is the number of
sinks $w_{l+1}, \dots, w_{l+k}$ in $E$, and for every $j\in \{1, \dots, k\}$, $n_j$
is the number of paths ending in the sink $w_{l+j}$, then this algebra isomorphism is  a $^\ast$-isomorphism.
\end{itemize}
\label{necessary}
\end{proposition}
\begin{proof} (i) Assume that $E$ is a no-exit graph. Then $L(E)$ is noetherian by Theorem \ref{noetherian_NE_theorem} and hereditary by \cite[Theorem 3.5]{AMP}. In this case, a result from \cite[Example 3, p. 235]{Stenstrom} states that the maximal and total left rings of quotients are equal. So, $Q(E)=\Qltot(L(E))=\Qlmax(L(E)).$ Thus $Q(E)$ is left self-injective since $\Qlmax(L(E))$ is. On the other hand, a left self-injective and left hereditary ring is semisimple (\cite[Theorem 7.52]{Lam}) and thus it is right self-injective as well. But a regular left and right self-injective ring is unit-regular by \cite[Theorem 9.29]{G}.

(ii) Following the proof of \cite[Theorem 3.8]{AAS2}, the basis of $L(E)$ can be described as follows. Let $\Lambda_i,$ $i=1,\ldots,l,$ be the set of paths ending in a fixed vertex $v_{m_i}$ of the cycle $c_i$ which do not contain the cycle itself, and let $\Lambda_j,$ $j=l+1,\ldots, l+k,$ be the set of paths ending in a sink $w_{l+j}.$ Let $\Lambda=\bigcup_{t=1}^{l+k} \Lambda_t.$ If $\lambda$ denotes the cardinality of $\Lambda,$ index the elements of $\Lambda$ as $p_s,$ $s=1,\ldots, \lambda.$ Finally, let
\[X = \{\, p_r c^z_t  p_s^*\, |\, z \in \Zset,\, r,s = 1,\ldots, \lambda,\,  t = 1,\ldots, l+k\,\}\]
where $c_t$ denotes $w_t$ for $t > l$ and $c_t^{-z}$ denotes $(c_t^*)^z$ for positive $z$ and
$t=1,\ldots, l.$
In \cite[Theorem 3.8]{AAS2} it is shown that the nonzero elements of $X$ constitute a basis of $L(E)$ and that $\phi$ is the isomorphism mapping a nonzero element $p_r c^z_t  p_s^*$ of $X$ for $t\leq l$ to $x^z e_{rs}$ where $e_{rs}$ is the standard matrix unit in appropriate matrix algebra $M_{m_i}(K[x,x^{-1}])$ for appropriate $m_i.$ For $t>l,$ $\phi$ maps a nonzero element  $p_r w_t p_s^*=p_rp_s^*$ of $X$ to the standard matrix unit $e_{rs}$ in the appropriate matrix algebra $M_{n_j}(K)$ for appropriate $n_j.$

The isomorphism $\phi$ is obtained as an $K$-algebra extension of this map of the basis elements. This map has to be $^\ast$-isomorphism then also because on the basis elements,
$\phi((p_r c^z_t  p_s^*)^*)=\phi(p_s c_t^{-z}p_r^*)=x^{-z}e_{sr}=(x^ze_{rs})^*=(\phi(p_r c^z_t  p_s^*))^*$ if $t\leq l.$ If $t>l$ the claim follows similarly.
\end{proof}

This proposition gives us that the involution of a noetherian Leavitt path algebra $L(E)$ corresponds to the conjugate transpose involution of the sum of matrix algebras $\left(\bigoplus_{i=1}^l M_{m_i}(K[x, x^{-1}])\right)\oplus
\left(\bigoplus_{j=1}^{k}M_{n_j}(K)\right)$ where the involution on $K[x,x^{-1}]$ is given by $q\mapsto \overline{q}$ for $q\in K$ and $x\mapsto x^{-1}.$

\begin{theorem} Let $E$ be a finite graph.
The following conditions are equivalent.
\begin{enumerate}
\item[(1)] $E$ is a no-exit graph.
\item[(6)] $L(E)$ contains no infinite set of orthogonal idempotents.
\item[(7)] $Q(E)$ is unit-regular.
\item[(8)] $Q(E)$ is directly finite (i.e. $xy=1$ implies that $yx=1$ for all $x$ and $y$).
\item[(9)] $L(E)$ is directly finite.
\item[(10)] $L(E)$ is finite (i.e. $x^*x=1$ implies that $xx^*=1$ for all $x$).
\item[(11)] $Q(E)$ is left and right self-injective.
\item[(12)] $Q(E)$ is semisimple.
\item[(13)] $L(E)$ has finite uniform dimension (as a left and as a right $L(E)$-module).
\item[(14)] The monoid of equivalence classes of finitely generated projectives $V(L(E))\cong V(Q(E))$ is cancellative.
\end{enumerate}
\label{equivalences1-14}
\end{theorem}

The gap in the numbering of the conditions (the first condition in Theorem \ref{equivalences1-14} is labeled by (1) and the second by (6)) indicates conditions (2)--(5) from Theorem \ref{noetherian_NE_theorem}. Any mention of (2)--(5) in the proof of Theorem \ref{equivalences1-14} refers to the conditions from Theorem \ref{noetherian_NE_theorem}.

\begin{proof}
(1) $\Rightarrow$ (6). Condition (1) implies that $L(E)$ is left and right noetherian by Theorem \ref{noetherian_NE_theorem}. But a noetherian ring contains no infinite set of orthogonal idempotents.

The implication (6) $\Rightarrow$ (1) is Proposition \ref{sufficient}.

The implication (1) $\Rightarrow$ (7) is Proposition \ref{necessary}.

The implication (7) $\Rightarrow$ (8) is \cite[Proposition 5.2]{G}.

The implications (8) $\Rightarrow$ (9) $\Rightarrow$ (10) hold by definition.

The implication (10) $\Rightarrow$ (1) is Proposition \ref{sufficient}.
Thus, conditions (1) -- (10) are equivalent.

The implication (1) $\Rightarrow$ (11) is Proposition \ref{necessary}.

The implication (11) $\Rightarrow$ (7) is \cite[Theorem 9.29]{G}.

(11) $\Leftrightarrow$ (12). One direction holds because a hereditary and self-injective ring is semisimple. Conversely, a semisimple ring is both hereditary and self-injective.

The equivalences (13) $\Leftrightarrow$ (2) and (3) follow from \cite[Theorem
7.58]{Lam}.

(14) $\Leftrightarrow$ (7). \cite[Theorem 2]{H} asserts that $Q(E)$ is unit-regular if and only if $V(Q(E))$ is cancellative. The isomorphism of monoids $V(Q(E))\cong V(L(E))$ is shown in \cite[Theorem 4.2]{AB} and \cite[Theorem 3.5]{AMP}.

Thus, conditions (1) -- (14) are equivalent.
\end{proof}

\begin{remark}
The authors are grateful to the referee for pointing out that the conditions (1)--(14) are also equivalent with the statement that $L(E)$ is a polynomial identity (PI) ring. Since a matrix ring over a commutative ring is a PI ring and a finite direct product of PI rings is a PI ring, condition (5) implies that $L(E)$ is a PI ring. Conversely, if $L(E)$ is a PI ring, then it is directly finite so the condition (9) holds.
\end{remark}

Theorem \ref{equivalences1-14} has the following corollary.
\begin{corollary}
If $E$ is a finite no-exit graph, then  $L(E)$ is Baer.
\label{Baer}
\end{corollary}
\begin{proof}
\cite[Theorem 7.55]{Lam} states that for a ring satisfying condition (6) from Theorem \ref{equivalences1-14}, the following conditions are equivalent: being Baer, being right Rickart, and being left Rickart. Since $L(E)$ is Rickart (as a semihereditary ring) and (6) is equivalent to (1), $L(E)$ is Baer.
\end{proof}

Now we turn to the question whether the involution of $L(E)$ extends to $Q(E).$ Recall that a regular ring with involution is said to be \emph{$^\ast$-regular} if the involution is proper (see \cite[Exercise 6A, \S 3]{Be}). Also recall \cite[Theorem 3.1]{Ara_Menal} stating that a $^\ast$-regular ring $R$ must necessarily be finite.

\begin{proposition} If the involution on $K$ is positive definite and if the involution $^\ast$ of $L(E)$ extends from $L(E)$ to $Q(E),$ then this extension is  positive definite and $Q(E)$ is a finite $^\ast$-regular ring.
\label{if_extends_star_regular}
\end{proposition}
\begin{proof}
Let us first show that the involution on $Q(E)$ is positive definite. Assume that for some $n\geq 1$ we have $\sum_{i=1}^n q_{i}^*q_{i}=0$ with $q_{i}\in Q(E)$. Write $p_{i}=q_{i}^*$ so that we have $\sum_{i=1}^n p_{i}p_{i}^*=0$. Assume that there is some $q_{i}$, say $q_{1}$, that is nonzero. In that case $p_{1}$ is nonzero too.

The ring $Q(E)=\Qltot(L(E))$ is a left ring of quotients of $L(E)$ and $L(E)$ is dense in $Q(E)$ as a left $L(E)$-submodule of $Q(E)$ ($L(E)$ embeds into $\Qltot(L(E))$ by \cite[Theorem 4.1]{Stenstrom}). Thus, applying \cite[Exercise 9, p. 284]{Lam} we can find $r\in L(E)$ such that $rp_{1}\neq 0$ and $rp_{i}\in L(E)$ for all $i$.

Now $\sum_{i=1}^n p_{i}p_{i}^*=0$ implies that $0=r(\sum_{i=1}^n p_{i}p_{i}^*)r^*=\sum_{i=1}^n (rp_{i})(rp_{i})^*=0$. But applying \cite[Proposition 2.4]{ARV} we have that $^\ast$ is positive definite in $L(E)$ if the involution on $K$ is positive definite. Therefore we have that $rp_{i}=0$ for all $i$, which is a contradiction with the fact that $rp_{1}\neq 0$. Therefore $q_{i}=0$ for all $i.$ This shows that the involution on $Q(E)$ is positive definite.

For a regular ring to be $^\ast$-regular, it is sufficient for $^\ast$ to be proper. Since $^\ast$ is positive definite on $Q(E),$ $^*$ is proper. Thus, $Q(E)$ is $^\ast$-regular. In addition, $Q(E)$ is finite by \cite[Theorem 3.1]{Ara_Menal}.
\end{proof}

Note that if $E$ is a finite and acyclic graph, $L(E)=Q(E)$ and so the involution extends trivially. This is because $L(E)$ of an acyclic graph is regular by \cite[Theorem 1]{Gene-Ranga}.  A regular ring is equal to its total left (and right) ring of quotients (\cite[Example 1, p. 235]{Stenstrom}). Thus,  $L(E)=Q(E)$ since $Q(E)$ is the total left ring of quotients of $L(E).$

The next example demonstrates that the involution does not always extend from $L(E)$ to $Q(E).$

\begin{example} Let $K$ be any field that has a positive definite involution (for example, complex numbers with conjugated complex involution). Let $E$ be the graph of a single vertex $v$ and two edges $e,f$.
$$\xymatrix{ {\bullet}^{v} \ar@(ul,dl) _{e} \ar@(dr,ur) _{f}   }$$
Then the involution does not extend from $L(E)$ to $Q(E).$
\end{example}
\begin{proof}
If we assume that the involution extends from $L(E)$ to $Q(E),$ $Q(E)$ would be $^\ast$-regular and finite by Proposition \ref{if_extends_star_regular}. But this is not the case since $e^*e=1$ and $ee^*=1-ff^*\neq 1$ because $ff^*\neq0$ (since otherwise $f=f1=ff^*f=0$). So $Q(E)$ is not finite. Hence the involution does not extend from $L(E)$ to $Q(E).$
\end{proof}

Any extension of the involution from $L(E)$ to $Q(E)$ has to be unique as the following proposition shows.
\begin{proposition} If the involution $^\ast$ of $L(E)$ extends from $L(E)$ to $Q(E),$ then this extension is unique.
\end{proposition}
\begin{proof}
Assume that there are two involutions extending the involution of $L(E)$ to $Q(E).$ Then their composition is a ring automorphism of $Q(E)$ that leaves $L(E)$ fixed. Consider the difference $f$ between this automorphism and the identity map. The map $f$ has $L(E)$ in the kernel and so $f$ factors to a map $\overline{f}: Q(E)/L(E)\rightarrow Q(E)$. We claim that $\overline{f}$ is zero since it maps a torsion module into a torsion-free module. Indeed, $Q(E)$ is torsion-free by \cite[Proposition 1.8, p. 198]{Stenstrom} with respect to the torsion theory that makes $Q(E)$ into the total left ring of quotients of $L(E)$ (more details on this torsion theory can be found in \cite{Stenstrom}). The module $Q(E)/L(E)$ is the cokernel of the monomorphism $L(E)\subseteq Q(E),$ the natural map from $L(E)$ to $Q(E)=\Qltot(L(E)).$  The cokernel $Q(E)/L(E)$ is a torsion module by \cite[Lemma 1.5, p. 196]{Stenstrom}.
Thus, $\overline{f}$ is a map from a torsion to a torsion-free module and so it has to be zero. Hence $f$ is zero as well and so the involution extends uniquely.
\end{proof}

The next result shows that further equivalences can be added to the list of those in Theorem \ref{equivalences1-14} in case when the involution on $K$ is positive definite.

Recall that a $^\ast$-ring $R$ is said to be {\em symmetric} if $1+x^*x$ is invertible for every $x.$ In this case, $R$ has a property that for every idempotent $a$ there is a projection (selfadjoint idempotent) $p$ such that $aR=pR$ (\cite[Exercise 7C, p. 9]{Be}).

\begin{theorem} Let $E$ be a finite graph.
If the involution on $K$ is positive definite, then (15)--(19) are equivalent to (1)--(14).
\begin{itemize}
\item[(15)] The involution $^\ast$ extends from $L(E)$ to $Q(E).$
\item[(16)] $Q(E)$ is $^\ast$-regular (for the involution inherited from $L(E))$.
\item[(17)] $Q(E)$ is finite (for the involution inherited from $L(E))$.
\item[(18)] $L(E)$ is extendible (i.e. $^\ast$ extends to $\Qlmax(L(E))$) and $Q(E)=\Qlmax(L(E))$.
\item[(19)] $Q(E)$ is a symmetric $^\ast$-ring (for the involution inherited from $L(E))$.
\end{itemize}
\label{equivalences15-19}
\end{theorem}

\begin{proof}
(5) $\Rightarrow$ (15). Let $\phi$ be the map from $L(E)$ onto $R=\left(\bigoplus_{i=1}^l M_{m_i}(K[x, x^{-1}])\right)\oplus$ $\left(\bigoplus_{j=1}^{k}M_{n_j}(K)\right)$ as in (5). This isomorphism induces the isomorphism (we call it also $\phi$) of the maximal left rings of quotients $Q(E)=\Qlmax(L(E))$ (see Proposition \ref{necessary})
onto $\Qlmax(R)=\left(\bigoplus_{i=1}^l M_{m_i}(K(x))\right)\oplus\left(\bigoplus_{j=1}^{k}M_{n_j}(K)\right).$ For an element $q$ of $Q(E)$ define the involution by $q^*=\phi^{-1}(\phi(q)^*).$ This is well defined by Proposition \ref{necessary}
and defines an involution on $Q(E)$ that extends the one on $L(E)$ also by Proposition \ref{necessary}.

The implication (15) $\Rightarrow$ (16) is Proposition \ref{if_extends_star_regular}.

The implication (16) $\Rightarrow$ (17) is \cite[Theorem 3.1]{Ara_Menal}.

The implication (17) $\Rightarrow$ (1) follows from Proposition \ref{sufficient} since if $Q(E)$ is finite, then $L(E)$ is finite as well.

(15) $\Leftrightarrow$ (18). Note that (1) implies that $Q(E)=\Qlmax(L(E))$ (this fact is in the proof of Proposition \ref{necessary}). Since (15) + (1) implies (18) and (15) implies (1), we have that (15) $\Rightarrow$ (18). Finally, (18) clearly implies (15).

(15) $\Rightarrow$ (19). A $^\ast$-regular ring with 2-proper ($x^*x+y^*y=0$ implies $x=y=0$) involution is symmetric (\cite[Exercise 9A p. 232]{Be}). Thus, a $^\ast$-regular ring with positive definite involution is symmetric. Condition (15) and Proposition \ref{if_extends_star_regular} imply that $Q(E)$ is a $^\ast$-regular ring with positive definite involution.

(19) $\Rightarrow$ (16). A $^\ast$-ring that is symmetric and regular is $^\ast$-regular (since symmetric and Rickart implies Rickart $^*$, and Rickart $^*$ and regular implies $^*$-regular, \cite[Exercises 6A, 7A p. 18]{Be}).
\end{proof}

\begin{remark} Conditions (1) -- (14) imply that $Q(E)$ is Baer (as any left or right self-injective and regular ring is Baer, see \cite[Corollary 7.53]{Lam}). Conditions (15) -- (18) imply that $Q(E)$ is Baer $^\ast$-ring (since $^\ast$-regular and Baer ring is regular Baer $^\ast$-ring).
\end{remark}

\begin{example} If $E$ is a finite no-exit graph, $Q(E)$ does not have to be symmetric if the involution of $K$ is not positive definite. For example, if $E$ is a loop (i.e. graph with one vertex and one edge), then $L_{\Zset_2}(E)=\Zset_2[x,x^{-1}]$ (\cite[Example 1.4]{Gene_Gonzalo1}) and $Q_{\Zset_2}(E)=\Zset_2(x)$. This algebra is not symmetric because $1+xx^*=1+xx^{-1}=1+1=0$ is not invertible.

Also, the involution on  $Q_{\Zset_2}(E)$ is not proper so $Q_{\Zset_2}(E)$ is not $^\ast$-regular. Thus we see that (1)--(14) hold but not (16), for example. So, this example shows that condition (15)--(18) are not necessarily equivalent with (1)--(14) if the involution of the field $K$ is not positive definite.
\end{example}

Theorem \ref{equivalences15-19} extends the class of known rings for which the Handelman's conjecture holds. Recall that the Handelman's conjecture (\cite[Problem 48, p. 380]{G}) is asking if every $^\ast$-regular ring is directly finite and if it is unit-regular. In \cite[Corollary 4.2]{ARV} it was shown that every  $^\ast$-regular Leavitt path algebra is unit-regular. From this result also follows that a $^\ast$-regular Leavitt path algebra is directly finite (since a unit-regular ring is directly finite by \cite[Proposition 5.2]{G}.)

Theorem \ref{equivalences15-19} asserts that the Handelman's conjecture holds for the class of regular algebras of Leavitt path algebras: in case when the involution on $K$ is positive definite and $Q(E)$ is equipped with the involution originating from $L(E),$ the regular algebra $Q(E)$ is $^\ast$-regular and both directly finite and unit-regular.

\section{$Q(E)$ as a ring of quotients}
\label{section_symmetric}

In this section, we explore the properties of $Q(E)$ as both one-sided and symmetric ring of quotients. Let us recall a few facts on rings of quotients first. Let $\ef_r$ be a right Gabriel filter (i.e. a set of right ideals of a ring $R$ that defines a hereditary torsion theory $\tau$, see \cite{Stenstrom} for more details). The right ring of quotients of $R$ with respect to $\ef_r$ (and $\tau$) is denoted by $R_{\ef_r}.$ This ring can be represented as $\dirlim_{I\in\ef_r}\homo(I, \frac{R}{\te(R)})$ where $\te(R)$ is the torsion submodule of $R$. Equivalently, $q$ is an element of $R_{\ef_r}$ if there is a right ideal $I\in \ef_r$ and a right module homomorphism $f:I\rightarrow \frac{R}{\te(R)}$ such that $f(x)=qx$ for every $x\in I.$ A left-sided version is defined similarly.

Now consider an involutive ring $R$. We define a left Gabriel filter $\ef_l$ and a right Gabriel filter $\ef_r$ to be {\em conjugated} if and only if $\ef_l^*=\ef_r$ (i.e. $I\in \ef_r$ if and only if $I^*=\{r^*\ |\ r\in I\}\in \ef_l).$

\begin{proposition}
Let $R$ be an involutive ring and $\ef_l$ and $\ef_r$ conjugated left and right Gabriel filters. If the involution extends to $R_{\ef_r}$ then $R_{\ef_r}$ is also a left ring of quotients and $_{\ef_l}R=R_{\ef_r}.$ Similarly, if it extends to $_{\ef_l}R,$ then $_{\ef_l}R$ is a right ring of quotients and $_{\ef_l}R=R_{\ef_r}.$
\label{right_is_symmetric}
\end{proposition}
\begin{proof}
Let $\tau_r=(\te_r,\f_r)$ and $\tau_l=(\te_l,\f_l)$ denote the torsion theories of right and left $R$-modules that $\ef_r$ and $\ef_l$ determine. Note that the involution maps the left torsion submodule $\te_l(R)$ onto the right torsion submodule $\te_r(R).$ This is because $r\in\te_r(R)$ if and only if $rI=0$ for some right ideal in $\ef_r.$ But then $I^*r^*=0$ and $I^*$ is in $\ef_l$ and so $r^*\in\te_l(R).$ Similarly, $r\in\te_l(R)$ implies that $r^*\in\te_r(R).$

Let now $q$ be in $R_{\ef_r}.$ Since the involution extends to $R_{\ef_r},$ $q^*$ is in $R_{\ef_r}$ as well. So, there is a right ideal $I\in\ef_r$ and a right $R$-homomorphism $f:I\rightarrow R/\te_r(R)$ such that $f(x)=q^*x.$ Then $I^*$ is a left ideal and the map $f^*:I^*\rightarrow R/\te_l(R)$ defined by $f^*(x)=f(x^*)^*$ is such that $f^*(x)=xq.$ Thus $q$ is in $_{\ef_l}R.$ The converse is proven similarly: if $q$ is in $_{\ef_l}R$ and $I$ a left ideal with a homomorphism $f:I\rightarrow R/\te_l(R)$ such that $f(x)=xq,$ then $I^*$ is a right ideal and $f^*:I^*\rightarrow R/\te_r(R)$ given by $f^*(x)=f(x^*)^*$ is such that $f^*(x)=q^*x.$ Thus $q^*$ is in $R_{\ef_r}.$ But then $q=(q^*)^*$ is in $R_{\ef_r}$ as well.
\end{proof}

If $\ef_l$ and $\ef_r$ are left and right Gabriel filters, the symmetric filter $_l\ef_r$ induced by $\ef_l$ and $\ef_r$ is defined to be the set of (two-sided) ideals of $R$ containing ideals of the form $IR+RJ$, where $I\in\ef_l$ and $J\in \ef_r$ (equivalently, the set of right ideals of $R\otimes_{\Zset}R^{op}$ containing ideals of the form $J\otimes R^{op}+R\otimes I$).
This defines a Gabriel filter by \cite[p. 100]{Ortega_thesis}. The corresponding torsion theory induced by $\tau_l$ and $\tau_r$ is denoted by  $_l\tau_r.$ If $M$ is an $R$-bimodule, $\te_l(M),$ $\te_r(M)$ and $_l\te_r(M)$ torsion submodules of $M$ for $\tau_l,$ $\tau_r$ and  $_l\tau_r$ respectively, then $_l\te_r(M)=\te_l(M)\cap\te_r(M).$

In \cite{Ortega_thesis} and \cite{Ortega_paper}, Ortega considers the symmetric ring of quotients $_{\ef_l}R_{\ef_r}$ (or $_lR_r$ for short) with respect to $_l\ef_r$ to be \[_lR_r=\dirlim_{K\in _l\ef_r}\;  \homo(K, \frac{R}{_l\te_r(R)})\]
where the homomorphisms in the formula are $R$-bimodule homomorphisms. Ortega shows that an equivalent approach can be obtained considering compatible pairs of homomorphisms. If $I\in\ef_l,$ $J\in\ef_r,$ $f:I\rightarrow\, \overline{R}$ and $g:J\rightarrow \overline{R}$ are homomorphisms with $\overline{R}=R/_l\te_r(R),$ then $(f,g)$ is a {\em compatible pair} if $f(i)j=ig(j)$ for all $i\in I,$ $j\in J.$ Define the equivalence relation by $(f,g)\sim(h,k)$ if and only if $f|_{I'}=h|_{I'}$ for some $I'\in \ef_l$ and $g|_{J'}=k|_{J'}$ for some $J'\in \ef_r.$ Then there is a bijective correspondence between elements of $_lR_r$ and the equivalence classes of compatible pairs of homomorphisms (\cite[Proposition 4.37]{Ortega_thesis} or \cite[Proposition 1.4]{Ortega_paper}) such that an element $q$ of $_lR_r$ can be represented by $(f,g)$ $f:I\rightarrow\overline{R}$ and $g:J\rightarrow\overline{R}$ if $f(i)=iq$ and $g(j)=qj$ for all $i\in I$ and $j\in J.$ In \cite{Ortega_thesis} and \cite{Ortega_paper}, Ortega further defines the symmetric module of quotients $_{\ef_l}M_{\ef_r}$ with respect to left and right Gabriel filters $\ef_l$ and $\ef_r$ of an $R$-bimodule $M$.

We prove the following property of symmetric rings of quotients of involutive rings.
\begin{proposition} Let $R$ be a ring with involution and $\ef_l$ and $\ef_r$ left and right Gabriel filters conjugated to each other.
\begin{enumerate}
\item[(i)] The involution extends to the symmetric ring of quotients $_lR_r.$

\item[(ii)] If $R$ is $\tau_r$-torsion-free ($\tau_l$-torsion-free) and the involution extends to $R_{\ef_r}$ ($_{\ef_l}R$)
then $R_{\ef_r}=\,_{\ef_l}R=\, _lR_r.$
\end{enumerate}
\label{involution_extends_to_symmetric}
\end{proposition}
\begin{proof}
First note that $_l\te_r(R)^*=(\te_l(R)\cap\te_r(R))^*=\te_r(R)\cap\te_l(R)=\,_l\te_r(R)$ if $\ef_l$ and $\ef_r$ are conjugated to each other.

(i) Let $(f,g)$ be a compatible pair representing the equivalence class of an element $q$ of the symmetric ring of quotients $_lR_r.$ Let $\overline{R}=R/_l\te_r(R),$ $I\in\ef_l,$ $J\in\ef_r,$ and $f:I\rightarrow\, \overline{R}$ and $g:J\rightarrow \overline{R}$ be homomorphisms with $f(i)=iq$ and $g(j)=qj.$  We can assume that $J=I^*$ (otherwise, we can replace the pair $(f,g)$ by an equivalent pair of homomorphisms defined on a left ideal generated by $I\cup J^*$ and a right ideal generated by $I^*\cup J$).

To define $q^*,$ consider a pair $(g^*, f^*)$ with $g^*:I\rightarrow \overline{R}$ and $f^*:I^*\rightarrow \overline{R}$ defined by $g^*(i)=g(i^*)^*$ and $f^*(j)=f(j^*)^*.$ It is a compatible pair since $g^*(i)j=g(i^*)^*j=(j^*g(i^*))^*=(f(j^*)i^*)^*=if(j^*)^*=if^*(j).$ This compatible pair represents the element $q^*$ since
$g^*(i)=g(i^*)^*=(qi^*)^*=iq^*$ and $f^*(j)=f(j^*)^*=(j^*q)^*=q^*j.$

(ii) Note that $ _lR_r$ embeds in $R_{\ef_r}$ by sending the class of a compatible pair $(f,g)$ to the equivalence class of $g.$ If the involution extends to $R_{\ef_r},$ then $R_{\ef_r}=\,_{\ef_l}R$ by Proposition \ref{right_is_symmetric}.

If $R$ is $\tau_r$-torsion-free, $\te_r(R)=0$ and so $\te_l(R)=(\te_r(R))^*=0$ and $_l\te_r(R)=\te_l(R)\cap\te_r(R)=0$ also.
We claim that $R_{\ef_r}$ embeds into $ _lR_r.$ To see this, let $q\in R_{\ef_r}$ be represented by $g:J\rightarrow R$ for some $J\in\ef_r.$ Since $q$ is also in $_{\ef_l}R,$ there is a left ideal $I$ and a map $f:I\rightarrow R$ that represents $q$ as an element of $_{\ef_l}R.$ Then $(f, g)$ is a compatible pair representing $q$ as an element of $_lR_r$ since
$f(i)j=(iq)j=i(qj)=ig(j).$
\end{proof}

If $\ef_l$ is the filter of dense left and $\ef_r$ a filter of dense right ideals, the symmetric ring of quotients induced by $\ef_l$ and $\ef_r$ is the maximal symmetric ring of quotients  $\Qmax(R)$ (introduced in \cite{Utumi}, studied in \cite{Lanning} and \cite{Ortega_paper_Qsimmax}). In particular, the classes of dense left and right ideals are conjugated so Proposition \ref{right_is_symmetric} and Proposition \ref{involution_extends_to_symmetric} can be applied to these filters.

The following lemma shows that perfect symmetric rings of quotients of involutive rings are also obtained via conjugated filters.
\begin{lemma} If $R$ is an involutive ring and $S$ a perfect symmetric ring of quotients with an injective localization map $q:R\rightarrow S$, then the involution extends to $S$ making $q$ a $^*$-homomorphism. With this involution, the left filter  $\ef_l=\{I | Sq(I)=S\}$ is conjugated to the right filter $\ef_r=\{J | q(J)S=S\}.$
\label{symmetric_perfect_have_involution}
\end{lemma}
\begin{proof} If  $S$ a perfect symmetric ring of quotients with localization map $q$, then $S$ is a symmetric ring of quotients with respect to the torsion theory induced by the left filter $\ef_l=\{I | Sq(I)=S\}$ and the right filter  $\ef_r=\{J | q(J)S=S\}$
by \cite[Theorem 4.1]{Lia_symmetric}. Note that if $I\in \ef_l,$ and $J\in \ef_r,$ then the left ideal $I+J^*$ is in $\ef_l$ and the right ideal $I^*+J$ is in $\ef_r.$ So, a compatible pair $(f,g)$ such that $f:I\rightarrow R$ and $g: J\rightarrow R$ can be exchanged by a compatible pair equivalent to it (call it $(f,g)$ again) defined on two ideals conjugated to each other. For $s\in S$ represented by such $(f,g),$ we can define $s^*\in S$ by representing it with the compatible pair $(g^*, f^*)$ defined in the same way as in the proof of Proposition \ref{involution_extends_to_symmetric}.

This definition makes $q$ a $^*$-homomorphism: for $r\in R$ the image $q(r)$ can be represented by a compatible pair $(R_r, L_r)$ where $L_r$ is the left and $R_r$ is the right multiplication by $r.$ Then the element $q(r)^*$ is represented by $(L_r^*, R_r^*)$ which is exactly $(R_{r^*}, L_{r^*})$ that represents $q(r^*).$ Thus $q(r^*)=q(r)^*$ for all $r\in R.$

With such involution on $S$, it is easy to see that $I\in \ef_l$ if and only if $I^*\in \ef_r.$ So, the filters are conjugated.
\end{proof}

In \cite{Lia_symmetric}, the total symmetric ring of quotients $\Qtot$ is defined as a symmetric version of the total one-sided rings of quotients $\Qrtot$ and $\Qltot.$ From Lemma \ref{symmetric_perfect_have_involution}, it follows that the involution extends to the total symmetric ring of quotients $\Qtot(R).$ Note that $R$ always embeds into $\Qtot(R)$ (\cite[Theorem 5.1]{Lia_symmetric}) so the assumptions of Lemma \ref{symmetric_perfect_have_involution} are satisfied. Also $R$ also embeds in $\Qrtot(R),$ $\Qltot(R)$ (see \cite[Theorem 4.1]{Stenstrom}) so Proposition \ref{right_is_symmetric} and Proposition \ref{involution_extends_to_symmetric} can be applied as well. Thus, the following corollary holds.

\begin{corollary} Let $R$ be an involutive ring.
\begin{enumerate}
\item[(i)] The involution extends to $\Qmax(R)$ and $\Qtot(R).$

\item[(ii)] If the involution extends to $\Qrmax(R)$ or $\Qlmax(R),$ then $\Qrmax(R)=\Qlmax(R)=\Qmax(R).$

\item[(iii)] If the involution extends to $\Qrtot(R)$ or $\Qltot(R),$ then $\Qrtot(R)=\Qltot(R)=\Qtot(R).$
\end{enumerate}
\label{quotients_of_involutive_ring}
\end{corollary}

Let us consider Leavitt path algebras now. We have seen that the equivalent conditions (1)--(14) imply that $Q(E)=\Qlmax(L(E))=\Qltot(L(E)).$ In case that (15)--(19) hold as well, the involution extends to $Q(E).$ Thus the following holds.

\begin{corollary}
If the involution on $K$ is positive definite, then conditions (1)--(19) are equivalent with
\begin{itemize}
\item[(20)] $Q(E)$ is a symmetric ring of quotients induced by conjugated filters.

\item[(21)] $Q(E)=\Qmax(L(E)).$

\item[(22)] $Q(E)=\Qtot(L(E)).$
\end{itemize}
\label{Q(E)_as_quotient}
\end{corollary}
\begin{proof}
Any of (20), (21) or (22) implies (15) by Proposition \ref{involution_extends_to_symmetric}.
Conversely, (15) implies (21) and (22) by Corollary \ref{quotients_of_involutive_ring}.  Finally, (21) or (22) trivially imply (20).
\end{proof}

If $E$ is a finite graph, $Q(E)$ is the total left ring of quotients of $L(E)$ even if it may not be equal to $\Qlmax(L(E))$ in case that $L(E)$ is not noetherian. However, since $L(E)$ is hereditary (thus semihereditary as well), the description of the total left ring of quotients from \cite[Theorem 12]{Lia_quotients} can be used to describe $Q(E)$ via $\Qlmax(L(E)).$ Thus,
\[Q(E)=\{q\in \Qlmax(L(E))\ |\ Iq\subseteq R\mbox{ for a left ideal }I\mbox{ with }\Qlmax(L(E))I=\Qlmax(L(E))\}.\]

\section{Further equivalences and $K_0$-theorem}
\label{section_K_0}

Rings with involution for which the involution extends to one-sided maximal rings of quotients have been studied in the past (e.g. \cite{Handelman}, \cite{Pyle}). For Leavitt path algebras with (1)--(22), the involution can be extended to the left maximal rings of quotients. This fact implies some further properties that we discuss in this section. In the following proposition and its proof, $\Qrmax$ stands for $\Qrmax(L(E)),$ $\Qlmax$ stands for $\Qlmax(L(E))$ and similar abbreviations are used for the total and classical rings of quotients of $L(E)$ as well.

\begin{proposition} If $E$ is a finite graph and $K$ a field with positive definite involution, then the following conditions are equivalent with (1)--(22).
\begin{itemize}
\item[(23)] $\Qrmax=\Qltot.$

\item[(24)] Every finitely generated nonsingular $L(E)$-module is projective.

\item[(25)] $\Qrmax\otimes_{L(E)}\Qrmax\cong\Qrmax$ via $q\otimes s\mapsto qs$ and $\Qrmax$ is flat as a right and left $R$-module.

\item[(26)] $\Qrmax=\Qltot=\Qrtot.$

\item[(27)] $M_n(L(E))$ is right strongly Baer (i.e. every complemented right ideal is generated by an idempotent) for every $n.$

\item[(28)] Every finitely generated $L(E)$-module $M$ is such that $M\otimes_{L(E)} Q(E)= M\otimes_{L(E)} \Qrmax.$

\item[(29)] $\Qrmax=\Qlcl.$
\end{itemize}
Moreover, these conditions imply that $M\otimes_{L(E)}\Qrmax$ is isomorphic to the injective envelope $E(M)$ for any nonsingular right $L(E)$-module $M.$ Also, the following rings of quotients are all equal to $Q(E)$
$\Qrmax=\Qlmax=\Qmax=\Qrtot=\Qltot=\Qtot=\Qrcl=\Qlcl.$
\label{extendible_LPAs}
\end{proposition}

\begin{proof}
First we prove (23) $\Rightarrow$ (24) $\Rightarrow$ (25) $\Rightarrow$ (26) $\Rightarrow$ (23). Then we show that these conditions are equivalent with (1)--(22). Finally, we demonstrate that (27)--(29) are equivalent to the rest.

Condition (23) implies that $\Qrmax$ is a perfect left ring of quotients of $L(E).$ Since $L(E)$ is (right) semihereditary, condition (24) follows by \cite[Corollary 7.4, p.259]{Stenstrom}.

Also by \cite[Corollary 7.4, p.259]{Stenstrom} (see the last sentence of it), (24) implies that $\Qrmax$ both a perfect left and a perfect right ring of quotients. Since such quotients have property (25) (see \cite[Theorem 4.1]{Lia_symmetric}), (24) implies (25).

Condition (25) implies that $\Qrmax$ is a perfect left and a perfect right ring of quotients. In that case, it is also a perfect symmetric ring of quotients. By Lemma \ref{symmetric_perfect_have_involution}, the involution extends to $\Qrmax.$ Hence $\Qrmax=\Qlmax=\Qmax$ by Corollary \ref{quotients_of_involutive_ring}. Since (25) also implies that $\Qrmax\subseteq\Qrtot$ and $\Qrmax\subseteq\Qltot,$ the equalities follow since $\Qrmax\subseteq\Qrtot\subseteq\Qrmax$ and $\Qrmax\subseteq\Qltot\subseteq\Qlmax=\Qrmax.$ So, (26) holds.

Condition (26) trivially implies (23).

Condition (26) implies (22). This is because $\Qltot=\Qrtot$ implies that $\Qltot=\Qrtot=\Qtot$ and this implies (22) because $Q(E)$ is always equal to $\Qltot.$ Conversely, if (22) holds, then (21) and (18) hold as well and so $\Qtot$ is equal to $\Qltot=\Qlmax,$ thus $\Qrtot$ as well, and $\Qtot$ is also equal to $\Qmax$ and $\Qrmax$ by part (ii) of Corollary \ref{quotients_of_involutive_ring}. Thus (26) holds.

\cite[Theorem 2.4]{Evans_all} states that (27) is equivalent to the statement that $L(E)$ is right semihereditary with $\Qrmax$ that is a perfect left and a perfect right ring of quotients. Since this last condition is precisely (25) and $L(E)$ is indeed right semihereditary, (25) and (27) are equivalent.

(23) $\Leftrightarrow$ (28). From $Q(E)=\Qltot,$ we have that (23) implies (28). Taking $M=L(E)$ in (28), we obtain (23).

Condition (13) implies that $L(E)$ is a semiprime left and right Goldie ring. This is because $L(E)$ is semiprime, left and right nonsingular and has finite uniform dimension if we assume (13), so it is a semiprime two-sided Goldie ring (\cite[Theorem 11.13]{Lam}). Thus, it is left and right Ore with $\Qlmax=\Qlcl$ and $\Qrmax=\Qrcl$ (\cite[Corollary 13.15]{Lam}). A left and right Ore ring has $\Qrcl=\Qlcl$  (\cite[Remark 10.17]{Lam}). Thus $\Qrmax=\Qlcl$ and (29) follows. Conversely, we show that (29) implies (24). Since $\Qlcl$ is a perfect left ring of quotients (\cite[Example p.230]{Stenstrom}), (29) implies that $\Qrmax$ is a perfect left ring of quotients. Since $L(E)$ is right semihereditary, (24) follows by \cite[Corollary 7.4]{Stenstrom}.

The last sentence of the proposition holds by \cite[Corollary 2.8, p. 248]{Stenstrom}.
\end{proof}

\begin{remark}
Note that the algebra $M_n(L(E))$ is isomorphic to the Leavitt path algebra
$L(M_n E)$ where $M_nE$ is the graph obtained by adding the oriented line of
length $n-1$ to every vertex of $E$ (see \cite[Definition 9.1 and Proposition
9.3]{Abrams_Tomforde}). Note that this gives us $M_n(Q(E))\cong Q(M_nE)$ as
well. This follows from the fact that $Q(E)=\Qltot(L(E))$ and that
$M_n(\Qltot(R))\cong \Qltot(M_n(R))$  for any ring $R$.
\end{remark}

We use Proposition \ref{extendible_LPAs} to obtain a specific description of the
inverse of the isomorphism $V(L(E))\rightarrow V(Q(E))$ of the monoids of
equivalence classes of finitely generated projectives over $L(E)$ and $Q(E).$ In
\cite[Theorem 3.5]{AMP}, it is shown that there is natural isomorphism between
$V(L(E))$ and an abelian monoid $M_E$ defined via the generators $\{a_v\ |\ v\in
E^0\}$ and subject to relations
\begin{center}
$a_v=\sum a_{r(e)}$ for all $e\in E^1$ with $v=s(e)$ and every $v\in E^0$ that emits edges
\end{center} if $E$ is row-finite. \cite[Theorem 3.1]{AB} proves that there is a canonical isomorphism between $M_E$ and $V(Q(E)).$ In particular, from the proof it follows that the isomorphism $\varphi: V(L(E))\rightarrow V(Q(E))$ is induced by the map  $P\mapsto P\otimes_{L(E)}Q(E).$ Our goal is to prove that the inverse of the isomorphism $\varphi$ is induced by $P\mapsto P\cap L(E)^n$ if $P$ is a finitely generated projective $Q(E)$-module that can be embedded in $Q(E)^n.$

The relation between $L(E)$ and $Q(E)$ in certain ways parallels the one between a finite von Neumann algebra (or a Baer $^\ast$-ring satisfying axioms as in \cite{Lia_dimension}) and its algebra of affiliated operators (or the regular ring of a Baer $^\ast$-ring). So, the proof of Lemma \ref{complements=summands} in what follows parallels that of \cite[Lemma 8]{Lia_dimension}. Also, the proof of Theorem \ref{K0_theorem} parallels the proof of \cite[Corollary 25]{Lia_dimension}. First, let us recall a few preliminary facts. Let $R$ be any ring and $M$ a nonsingular $R$-module. By \cite[Corollary 7.44']{Lam}, there is a one-to-one correspondence between closed submodules of $M$ and those of $E(M)$ (for the definition of closed submodules, see \cite[Definition 7.31]{Lam}). By the remark after the proof of \cite[Corollary 7.44']{Lam}, the closed submodules of $E(M)$ are precisely the direct summands of $E(M).$ Moreover, if $N$ is a closed module of $M$, then its closure in $E(M)$ is a copy of the injective envelope $E(N).$ By \cite[Proposition 7.44]{Lam}, a submodule $N$ of $M$ is closed if and only if $N$ is a complement in $M$. This gives us a one-to-one correspondence
 \[\{\mbox{complements in }M\} \longleftrightarrow\{\mbox{direct summands of }E(M)\}\] given by
$N \mapsto$ the closure of $N$ in $E(M)$ that is equal to a copy of $E(N).$ The inverse map is given by $L\mapsto L\cap M$ (see the proof of \cite[Corollary 7.44']{Lam}). This gives us the following lemma.

\begin{lemma} Let $K$ be a field with positive definite involution, $L(E)$ be a noetherian Leavitt path algebra and let $P$ be a right $L(E)$-submodule of $L(E)^n$ for some non-negative $n$.
The following conditions are equivalent:
\begin{enumerate}
\item[(i)] $P$ is a complement in $L(E)^n.$

\item[(ii)] $P$ is closed in $L(E)^n.$

\item[(iii)] There is an idempotent $p\in M_n(L(E))$ such that $P=$ im $p.$

\item[(iv)] $P$ is a direct summand of $L(E)^n.$
\end{enumerate}
\label{complements=summands}
\end{lemma}
\begin{proof} Since $L(E)$ is a nonsingular ring (\cite[Proposition 4.1]{Mercedes_quotients}), $L(E)^n$ is a nonsingular module. Thus, (i) and (ii) are equivalent as noted in the paragraph preceding the lemma. It is also clear that (iii) and (iv) are equivalent.

Since a direct summand is a complement, (iv) implies (i). Finally, we show that (ii) implies (iv). Let $P$ be a closed submodule of  $L(E)^n$. Then $L(E)^n/P$ is nonsingular by \cite[Theorem 7.28]{Lam}. Thus, $L(E)^n/P$ is finitely generated nonsingular module. Then  $L(E)^n/P$ is projective by condition (24) of Proposition \ref{extendible_LPAs}. So, the embedding of $P$ into $L(E)^n$ splits and thus $P$ is a direct summand of $L(E)^n$.
\end{proof}

Now we can prove the following theorem.

\begin{theorem} Let $K$ be a field with positive definite involution and $L(E)$ a noetherian Leavitt path algebra.
\begin{enumerate}
\item[(i)] For every finitely generated nonsingular (equivalently projective) $L(E)$-module $P$, there is
a one-to-one correspondence \[\{\mbox{direct summands of }P\}
\longleftrightarrow\{\mbox{direct summands of }E(P)=P\otimes_{L(E)}Q(E)\}\] given by
$N \mapsto$ $N\otimes_{L(E)}Q(E) = E(N).$ The inverse map is given by
$L\mapsto L\cap P.$

\item[(ii)] The isomorphism $\varphi: V(L(E))\rightarrow V(Q(E))$ induced by the map  $P\mapsto P\otimes_{L(E)}Q(E)$ has the inverse induced by $P\mapsto P\cap L(E)^n$ if $P$ is a finitely generated projective $Q(E)$-module that can be embedded in $Q(E)^n.$
\end{enumerate}
\label{K0_theorem}
\end{theorem}
\begin{proof}
(i) follows directly from Lemma \ref{complements=summands}, the paragraph preceding it and the fact that $P\otimes_{L(E)}Q(E)$ is the injective envelope of $P$ for a nonsingular $P$ (see the last sentence of Proposition \ref{extendible_LPAs}). (ii) follows directly from (i).
\end{proof}

\section{Isomorphism conjecture}
\label{section_isomorphism_conjecture}

A well-known theorem by Gardner \cite[Theorem B]{Gardner} shows that if two $C^*$-algebras are isomorphic as algebras, then they are isomorphic as $^\ast$-algebras. With this in mind, in \cite[p. 20]{Abrams_Tomforde}, G. Abrams and M. Tomforde posed several isomorphism conjectures.

\begin{C1}{\rm If  $L_{\Cset}(E)\cong L_{\Cset}(F)$ as algebras, then $L_{\Cset}(E)\cong L_{\Cset}(F)$
as $^\ast$-algebras.}
\end{C1}

\begin{C2}{\rm If $L_{\Cset}(E)\cong L_{\Cset}(F)$ as algebras, then $C^*(E)\cong C^*(F)$ as $^\ast$-algebras.}
\end{C2}

\begin{IC}{\rm If
$L_{\Cset}(E)\cong L_{\Cset}(F)$ as rings, then $C^*(E)\cong C^*(F)$
as $^\ast$-algebras.}
\end{IC}

It is known (see \cite[Corollary 4.5]{Abrams_Tomforde}) that if $L_{\Cset}(E)\cong L_{\Cset}(F)$ as $^\ast$-algebras, then $C^*(E)\cong C^*(F)$ as $^\ast$-algebras. Thus, (C1) implies (C2).

For these conjectures, it is assumed that the involution on $\Cset$ is fixed to be the complex-conjugate involution \cite[Definition 2.1]{Abrams_Tomforde}. If the involution in (C1) is not fixed, the conjecture trivially fails as the next example shows.

\begin{example}
Let $id$ denote the identity involution on $\Cset$ and let $\hskip.1cm^{\overline{\hskip0.2cm}}$ denote the complex-conjugate involution. Since $(\Cset, id)\cong(\Cset,  \hskip.1cm^{\overline{\hskip0.2cm}})$ as fields, $L_{(\Cset, id)}(E)\cong L_{(\Cset,  \hskip.1cm^{\overline{\hskip0.2cm}})}(E)$ as algebras. However, $L_{(\Cset, id)}(E)$ and $L_{(\Cset,  \hskip.1cm^{\overline{\hskip0.2cm}})}(E)$ are not isomorphic as $^\ast$-algebras: if $f$ is a $^\ast$-isomorphism then $i=if(1)=f(i)=f(i^*)=f(i)^*=(f(1)i)^* = i^*f(1)^*=-if(1)=-i$. A contradiction.
\end{example}

Working with general Leavitt path algebras does not imply any specific ties to the field of complex numbers equipped with the complex-conjugated involution. Thus, we generalize (C1) to a conjecture we call the Generalized Isomorphism Conjecture. We also add a stronger version of it -- the Generalized Strong Isomorphism Conjecture.

\begin{GIC}{\rm Let $K$ be a field with (fixed) involution. If  $L_{K}(E)\cong L_{K}(F)$ as algebras, then $L_{K}(E)\cong L_{K}(F)$ as $^\ast$-algebras.}
\end{GIC}

\begin{GSIC}{\rm Let $K$ be a field with (fixed) involution. If  $L_{K}(E)\cong L_{K}(F)$ as rings, then $L_{K}(E)\cong L_{K}(F)$ as $^\ast$-algebras.}
\end{GSIC}

Note that (GIC) implies (C1) and (C2) and (GSIC) implies (IC) (by \cite[Corollary 4.5]{Abrams_Tomforde}) and (GIC).

As we have seen in the example above, (GIC) and (GSIC) fail if the involution on $K$ is not fixed. Also, it is easy to see that a fixed isomorphism of Leavitt path algebras does not have to be a $^\ast$-isomorphism.

\begin{example}
Let $K$ be any field of characteristic different from 2 with any involution. Consider the Leavitt path algebra $L_K(E)$ over the 2-line graph.
$$\xymatrix{ ^{v}{\bullet} \ar[r]^e  &{\bullet}^{w} }$$
$L_K(E)$ is $^\ast$-isomorphic to the algebra $M_2(K)$ of $2\times 2$ matrices over $K$ via the isomorphism $\phi$ of Proposition \ref{necessary}. Now consider any invertible but non-unitary matrix from $M_2(K).$ For example, we can take $A=\left[\begin{array}{cc}2 & 0\\ 0 & 1 \end{array}\right]$ since char$(K)\neq 2.$ The map $f_A(X)=AXA^{-1}$ is an automorphism of $M_2(K)$ that is not a $^\ast$-isomorphism. Then $\phi^{-1}f_A\phi$ is an automorphism of $L_K(E)$ that is not a $^\ast$-isomorphism.
\end{example}

In \cite{Abrams_Tomforde}, it is shown that (C1) and (IC) hold when $E$ and $F$ are acyclic graphs \cite[Proposition 7.4]{Abrams_Tomforde}. It is also shown that (IC) holds if $E$ and $F$ are row-finite cofinal graphs with at least one cycle and such that every cycle has an exit \cite[Proposition 8.5]{Abrams_Tomforde}.

We give a positive answer for (GSIC) (thus for (GIC) and (IC) as well) for the Leavitt path algebras of finite no-exit graphs.

\begin{theorem}\label{isomorphismconjecture}
Let $E$ and $F$ be finite no-exit graphs. The following conditions are equivalent.
\begin{enumerate}
\item[{\rm (i)}] $L(E)\cong L(F)$ (as rings).
\item[{\rm (ii)}] $L(E)\cong L(F)$ (as algebras).
\item[{\rm (iii)}] $L(E)\cong L(F)$ (as $^\ast$-algebras).
\end{enumerate}
\end{theorem}
\begin{proof}
Suppose that $L(E)\cong L(F)$ as rings. By Theorem \ref{noetherian_NE_theorem}, there exist integers $l$, $k$, $l'$, $k'$, $m_i$, $n_j$, $m'_i$, $n'_j$ such that $L(E)$ is isomorphic to $R=\left(\bigoplus_{i=1}^l M_{m_i}(K[x, x^{-1}])\right)\oplus
\left(\bigoplus_{j=1}^k M_{n_j}(K)\right)$ and $L(F)$ isomorphic to $S=\left(\bigoplus_{i=1}^{l'} M_{m'_i}(K[x, x^{-1}])\right)\oplus
\left(\bigoplus_{j=1}^{k'} M_{n'_j}(K)\right)$ as algebras. By Proposition \ref{necessary}, these isomorphisms are $^\ast$-isomorphisms. Denote the $^*$-isomorphism $L(E)\to R$ by $\phi_E$ and the $^*$-isomorphism $L(F)\to S$ by $\phi_F.$ The ring isomorphism $L(E)\cong L(F)$ induces the ring isomorphism $R\cong S.$ Denote this last isomorphism by $\Phi.$

We show that $l=l'$, $k=k'$ and the sizes of the matrices match after reordering. Thus, there is a $^\ast$-isomorphism $f$ between $R$ and $S$ and the $^\ast$-isomorphism $\phi_F^{-1}f\phi_E$ from $L(E)$ onto $L(F)$.

First assume that $l>0$ and consider the ideal $I:=M_{m_1}(K[x,x^{-1}])\oplus 0 \oplus \dots \oplus 0.$
There exist ideals $I_i$ of $K[x,x^{-1}]$ and $J_j$ of $K$ such that $\Phi(I)=\left(\bigoplus_{i=1}^{l'} M_{m'_i}(I_i)\right)\oplus\left(\bigoplus_{j=1}^{k'} M_{n'_j}(J_j)\right)$.

In particular we have the ring isomorphism $$M_{m_1}(K[x,x^{-1}])\cong \left(\bigoplus_{i=1}^{l'} M_{m'_i}(I_i)\right)\oplus \left(\bigoplus_{j=1}^{k'} M_{n'_j}(J_j)\right),$$ which yields, by taking centers, the ring isomorphism $$K[x,x^{-1}]{\rm Id}_{m_1}\cong \left(\bigoplus_{i=1}^{l'} Z(I_i) {\rm Id}_{m'_i}\right)\oplus
\left(\bigoplus_{j=1}^{k'} Z(J_j) {\rm Id}_{n'_j}\right),$$ where for any $m\in {\mathbb N}$, ${\rm Id}_m$ denotes the identity
matrix of size $m$. But since $Z(I_i)=I_i,$ $Z(J_j)=J_j$, and $K[x,x^{-1}]$ does not have zero divisors, the equation
above is only possible if all but one $I_i$ or $J_j$ are zero. Assume $I_i\neq 0$ for some $i$. Then $${\mathbb M}_{m_1}(K[x,x^{-1}])\cong I\cong
\Phi(I)=0\oplus \dots\oplus 0 \oplus {\mathbb M}_{m'_i}(I_i) \oplus 0 \dots \oplus 0\cong {\mathbb M}_{m'_i}(I_i).$$

Apply \cite[Exercise 14, p. 480]{Lam} to get that $m_1=m'_i$ and that $K[x,x^{-1}]\cong I_i$. In particular, $I_i$
is a unital ring. Since $I_i$ consists of polynomials, a degree argument shows that the only identity element possible for $I_i$ is $1\in K[x,x^{-1}]$. So $I_i$ must equal $K[x,x^{-1}]$.

Similarly we can see that if $J_j\neq 0$, then we would have that $K[x,x^{-1}]\cong J_j.$ This is impossible since $J_{j}$ is either $0$ or $K$, so this case cannot happen.

Then, we have the following equality: $$\Phi(I)=0\oplus \dots\oplus 0 \oplus
{\mathbb M}_{m_1}(K[x,x^{-1}]) \oplus 0 \dots \oplus 0.$$ So we can mod out both $I$ and $\Phi(I)$ in the isomorphism $\Phi$ to obtain an induced isomorphism in the quotients, therefore completely removing one Laurent polynomial-type matrix component on each side.

This, in particular, shows that $l'>0.$ In the same way, by using $\Phi^{-1},$ we obtain that $l'>0$ implies $l>0$ as well. Therefore we have $l>0$ if and only if $l'>0$. By following a descending process as above, this shows that $l=l'$ and that there exists a permutation $\sigma$ such that $\sigma(m_{i})=m'_i$.

Furthermore, after conveniently removing all the matrices over Laurent polynomials we are left with a ring isomorphism

$$\overline{\Phi}:\left(\bigoplus_{j=1}^k M_{n_j}(K)\right) \to \left(\bigoplus_{j=1}^{k'} M_{n'_j}(K)\right).$$

In this situation we can apply Wedderburn-Artin Theorem to readily have that $k=k'$ and that there exists a permutation $\tau$ such that $\tau(n_j)=n'_j$.

All this, together with Proposition \ref{necessary}, shows that (i) $\Rightarrow$ (iii). The implications (iii) $\Rightarrow$ (ii) $\Rightarrow$ (i) are trivial.
\end{proof}

We provide here an affirmative answer to both (GSIC) and (IC) for the class of Leavitt path algebras considered in this paper.

\begin{corollary}\label{gsic_holds} (GSIC) holds for the class of noetherian Leavitt path algebras. In particular, (IC) holds for the class of finite no-exit graphs.
\end{corollary}
\begin{proof}
Apply Theorem \ref{isomorphismconjecture} and \cite[Corollary 4.5]{Abrams_Tomforde}.
\end{proof}

Finally, it is interesting to note the following. On one hand, the Leavitt path algebras over finite no-exit graphs satisfy (IC) by Corollary \ref{gsic_holds}. On the other hand, Leavitt path algebras over row-finite and cofinal graphs in which every cycle has an exit also satisfy (IC) by \cite[Proposition 8.5]{Abrams_Tomforde} (the assumption that there has to be at least one cycle in \cite[Proposition 8.5]{Abrams_Tomforde} can be dropped since acyclic graphs also satisfy (IC)). This gives us that the Leavitt path algebras over the finite and cofinal graphs on the two opposite sides of the spectrum (either no exits at all or exits from every cycle) both satisfy (IC). This gives reasons for hope that (IC) may hold for Leavitt path algebras of graphs in between these two extreme cases.

redpajama_set_name: RedPajamaArXiv


text:
Knowledge of Language é um livro de linguística escrito pelo linguista estadunidense Noam Chomsky, publicada originalmente em 1986. Nesta obra, Chomsky trata de tópicos da filosofia da linguagem e da filosofia da mente. Argumenta-se que o estudo de estruturas linguísticas fornece uma visão sobre o funcionamento da mente humana.

Livros de linguística
Livros de Noam Chomsky
Livros de 1986

redpajama_set_name: RedPajamaWikipedia


text:
Q: nested vue instance inside vue instance I want to put vue instance inside a vue instance so i can use the the dom element as my template and not write the template in the javascript code
ie
    { template : 'template for the component' }
example :
main.js
var v2 = new Vue({
  el: '#child',
  data : {
    msg1 : 'child msg'
  }
});

Vue.component(v2);

var v = new Vue({
  el: '#parent',
  data : {
    msg : 'parent msg'
  }
});

index.html

{{msg}}

            {{msg1}}

as you can see i dont use template i use the el property so my template come from the html on the page the element #parent and #child in this case also as you can see i dont give a name to the component and i just register the vue instance as a component Vue.component(v2);
and not Vue.component('child',v2); for example
now this is working and everything is good but there is no documentation
to this all the examples show how you can register a component with a template to a vue instance
because there is no documentation I am a afraid that in the next version of vue may be it will not work any more
someone know if this is the case or not

redpajama_set_name: RedPajamaStackExchange


text:
Topic: Doris Day, Dies at 97
https://www.bbc.com/news/entertainment-arts-48257670
I saw this and unfortunately it's one of those incidents of not even realising she was still alive (heretofore). Her name was really famous and everyone knows Que Sera Sera -- don't think I've ever seen one of her films though.
Deeply saddened by this.
I was reading only yesterday that she was Mike Nichols' first choice to play Mrs. Robinson in THE GRADUATE. That would have been something to see—and probably won her an Oscar!
...she was Mike Nichols' first choice to play Mrs. Robinson in THE GRADUATE.
That would've been amazing!
Very saddened by this news. I just recently watched one my favorite movies that she starred in: The Man Who Knew Too Much.
RIP Ms. Day.
Rick Senger
Obviously a talented singer but also an underrated actress who rarely got serious roles thanks to wildly successful vehicles like PILLOW TALK. Look no further than the aforementioned THE MAN WHO KNEW TOO MUCH where her slow build up of ultimately unbridled yet restrained heartache / agita at the reveal of her son's kidnapping (due to Jimmy Stewart giving her a tranquilizer) was arguably the most powerful moment in a pretty fair movie. She was similarly effective and moving in the underrated (and deceptively titled) thriller MIDNIGHT LACE.
Apparently no funeral, memorial or marker per her wishes.

redpajama_set_name: RedPajamaCommonCrawl


text:
Double M Resources, LLC
About Double M Resources
Double M Resources is an executive search firm specializing in placement of professionals with analytical and information technology expertise. MM provides professional talent for permanent positions in the healthcare, credit, financial, insurance, marketing, opinion research, pharmaceutical, biotechnology and other industries.
Our approach is to:
Assess the skills, experience and compensation requirements of the job seekers.
Match qualified candidates with employers.
Schedule interviews with the hiring manager.
After placement, follow up with both parties to ensure complete satisfaction.
Mindy Marty is the President and CEO of Double M Resources.
Mindy has specialized in recruiting professionals in the areas of statistical analysis and information technology since 1997.
Mindy holds a Bachelor of Science from Illinois State University and graduated Cum Laude with Departmental Honors.
© Double M Resources, LLC 2023