With festival season on it's way, I thought a little boho-inspo is appropriate to send your way. Whether you're going to Coachella or any spring festival in your city, I'll be doing a little series to get you in the mood for those fun, free-spirited days.
This series is all inspired by Coachella,the biggest arts/music festival there is in California. I'm sure you've seen plenty of your favorite bloggers take on Coachella in their ripped jeans, flowy kimonos, fringe tops, mirrored shades that include dreamy desert backdrops, drinks to raise, and of course, live music that powers you all day. How does that sound to you?
I still have yet to attend Coachella, but you know, I'm ready to bring Coachella to me as Vegas has got the desert and almost everything else I'll ever need. Mostly it's the spirit and the style that you'll need, which is easy enough, becoming that bohemian goddess. Tip: Think like a goddess, to become one!
You may wonder, what makes a goddess festival ready? Well, every goddess has got to have a crown whether it's a flower crown, a jewelry head piece or a cool floppy hat. Mine is a DIY crown I made (I admit I had some help) from fake red roses (my favorite flowers) I have bought a while ago. They, for sure, came in useful for this look and I'm hopping this won't be the last time I'm wearing them.
Because it's always sunny where there's a music festival, you've got to have some cool shades. Anything will do as long as you're rocking them while enjoying yourself with your friends and having an unforgettable time.
Dressing up for Coachella is the most fun and one of the best things about any festival, so let your imagination run wild. For me it was all about the dress, it being flowy, cute with loose sleeves. Do you like it? I see it as a total festival favorite!
Accessorize to the max with lots of jewelry whether that be rings, bracelets, necklaces or all of the above. You can never go wrong. I even added my favorite gold/silver temporary tattoos, which I like to wear in the spring. If you're like me, never wanting to get a real tattoo, these are the next best option.
Will you be going to Coachella or any music festival this year?
Remember I'll be back with more spring festival inspiration, so follow my blog to experience it all with me. Also, don't forget to share this post with any of your bohemian friends, wild at heart and free at spirit.
aww thanks! ❤ Happy weekend!
Glad the hike was worth it :) Yay!!!
I loved the photos, I'm following your blog! Loved it! I'm also seeing old posts, ours are awesome!
Beautiful pictures! I love the boho vibe of this look.. perfect for Coachella!
The place just adds to the overall ambiance of the pictures. I love that dress. But is coachella an American thing only?
Thanks. Yes Coachella is a huge music festival at California, but there are all kinds of spring festivals happening all season long.
I totally love that gorgeous Coachella outfit! =) So pretty!
Wow, Amazing background and lovely dress!
Здравей, Ради! I absolutely love this kind of posts, they always inspire me for the upcoming festivals! I don't have any festival plans yet, but who knows? Your dress and crown make you look so boho! You are such an inspiration, Radi! Пожелавам ти приятен ден! До скоро!
Wonderful photos! Like it a lot!
This look is amazing! It totally screams coachella!!
Yesss, I've got the style down. Now I've got to find tickets to Coachella.
Thanks a lot. Happy day!
I love your dress. The shoes and bag are so nice too. Yoour choice for the place where to make the photos is fantastic, the color contrast is awesome.
As you say, Vegas has got the desert and so now you can bring the outfits to it! :) I love this look on you, especially the floral crown.
Ohhh yes, I've got the best of both worlds. The floral crown is a diy, happy it turned out so good.
PS: Thank you for your lovely comment on my blog, dear!
I'm thrilled that you like it Jessi. Really hoping to attend it one of these days. I hope you have fun.
Loving the coachella/ festival outfit inspo here babe! I love that cutout shoulders will still be making an appearance this Spring!
I love your dress babe and the gorgeous location of this shoot, wish I were going to Coachella, hopefully next year!
Thanks so much Whitney. I feel the same way. Maybe we should go together?!
This dreamy boho look is so perfect for Coachella Radi! That's incredible that you made the flower crown yourself- it's beautiful :) You should sell them!!!
Thanks Rachel for this comment and great idea. Have a great weekend!
Love this festival look. Your diy flower crown is beautiful.
Wow! Your style and this location are out of this world! Super Gorgeous!!!
Happy you think so. Thank you!
Love your photos and your dress!!
Ohh that's super sweet. It's all thanks to the help and great work of my photographer.
I loved the photos! The look is awesome!
Nice look:)and beautiful flower crown!
Beautiful dress and wonderful pictures!
I'm obsessed with your dress, Rad. You look stunning and flawless! You are such an inspiration!
super fresh and stylish :D nice combo!


Bayside Retreat sits high over the water with an easy wooden staircase leading down to the protected pier on Castle Rock Lake (2nd largest inland WI lake, 13,600 acres). This new home sits on a 1.5 acre parcel surrounded by tall pines … grill outdoors on the deck overlooking the bay … or sit around the bricked campfire roasting marshmallows and gazing at stars. Pontoon boat rental within 5 minutes drive, and convenient launch just across the bay. Great room with cathedral ceiling and hardwood flooring … ping-pong or air-hockey in lower level. Conservancy paths for hiking along the bay. Enjoy your vacation retreat … with all the modern conveniences you expect. This home will easily accommodate your family of 11 … with three spacious bedrooms (each with TV) and sleeping for one in the loft overlooking the great room. Boating, dining, hiking, fishing, golfing, beaches, ATV and snowmobiling, canoeing, horseback riding, nature and more! Only 35 minutes to the WI Dells!


text:
Gryphon's Gate by Spirithunter
Fiction » Fantasy Rated: K+, English, Fantasy & Adventure, Words: 3k+, Follows: 2, Published: 8/22/2007 Updated: 8/26/2007}
3 Chapter 2: Dragonflight
Well, I guess this is something I will continue normally. Now, if I can just finish it...
Have fun, and I hope this is a satisfactory chappie. I just finished it here at my relatives' house in Canada.
1-Dragonflight
Oregon did indeed have a law decreeing that a helmet must be worn when flying—most states did. In fact, most countries did. But the law was not what made Diana Lane decide to wear a helmet. No, she wore one to keep the roaring wind out of her ears. It also helped keep her long light brown hair out of her face. She also wore ski goggles to keep her brown eyes safe from the wind and any other objects in it.
She rode on the dragon's shoulders, just in front of its great wings. As soon as she was fourteen, old enough to legally fly (though not too much, too high nor too long), she had gotten a custom-made saddle as blankets were not enough to ease her aching legs. It was a light saddle as a dragon such as this could not withstand much more weight than that of a fit human. Besides, with anything else, Diana's legs ached after too long. This saddle allowed her to ride for hours without getting sore. She had not yet had a chance to ride for hours, though. Usually she flew to school, or north to Portland. Diana's longest flight session had barely been about an hour when she was flying for fun. The black dragon was an agile thing; he had pulled of barrel rolls and loops and corkscrews and dives. Diana's saddle strap and strong legs had successfully kept her on his back.
It was the beginning of August, and there was an entire month left before school began once again. Diana sighed. Summer vacation was too long, even if there were plenty of things to do, because at some point she got bored of those things. Even the thrill of flight got old after too long. She'd been doing it for at least a year. That wasn't to say she found it particularly boring, though. Right now she was going to Portland for the Saturday Market, a sprawling weekend market of stands and tents filled with bowls, glasswork, jewelry, clothing, even the less conventional things such as longbows and decorative knives. Diana and the dragon were soaring over the Willamette River, searching the waterfront for the market. She had brought her allowance and babysitting money from the past two months, which she had been saving for this very purpose.
They landed in a clear patch of grass. Diana dismounted and put her flying gear in her backpack. While she went shopping, Caelum would be given the freedom to fly around and do whatever he felt like. Diana had put a pager-like device in the saddle, and she could call it with her cell phone to alert the dragon when he would be needed. Then he would fly back and pick her up.
She hugged the black dragon around the neck. "Have fun, Caelum," she said, and stepped back as he leaped from the ground with a great flap of his wings. As he had no forearms—dragons only had four limbs, whether both sets were legs, feet, or one of each, as he did—Caelum had great, powerful hindlimbs and four-fingered wings with dark gray membrane and a claw-like thumb on the front. His wings were long and relatively narrow. He had perhaps a sixteen-foot wingspan, longer if anything.
Soon the dragon was airborne and Diana was happily making her way among the stalls of the Saturday Market.
Despite how much she wished to have a thousand dollars to spend on everything, she only had fifty. She ended up spending it on a black tee with small stitched dragon patterns around the collar and sleeves, twin necklaces of blue beads and silver rings, one each for her and for Caelum, and a silver hair twist to wrap around a thin tress of hair. It came with three charms of her choice, so she chose a feather, a crescent moon, and since they were out of dragon charms, a gryphon.
As far as anyone knew, gryphons were still mythical. Besides, it was impossible for any creature in this world to have an anatomy like that—six limbs, and on top of that gryphons were like a blend of two completely different animals, bird and cat. Not that anyone cared. Even the biologists were waiting for the day they saw one. So far every other creature discovered had pertained to the laws of nature—four-limbed dragons not quite big enough for their flight to be impossible, unicorns small with no unnatural body parts, phoenixes that were large, scarlet, exotic, fire-loving and long-living, though they did not actually burst into flame at death. Instead their mates found or made a fire and cremated the dead. Kirin looked like unicorns but larger, with an antler-like horn instead of the spire-like kind, and with rich fur, usually white, gray-brown, or black, and sometimes stained green by the grasses they rolled and played and slept in.
There were also feyhounds, which were like a cross between German shepherds and golden retrievers in build, and silver or gold in color. The main difference between them and dogs was that the hounds were much more serene and had a pair of gray-brown branching horns sweeping out from behind their eyes and just above their long, tapering ears. Why they had been named feyhounds, no one knew, but they were popular pets all the same. Diana used to own one, a golden creature named Patticus by her young brother in his effort to say 'Pegasus', but a few years ago he died.
From her backpack, Diana pulled out her cell phone, dialing a number. Several miles away, Caelum heard the familiar ring and took flight, saying goodbye to his friends. In a few minutes, he'd landed in the field before her and Diana ran toward him, already dressed in her flying gear.
"Hey, Cae," she greeted. Caelum nodded, a small, quick movement, as was his way to greet others. "I got you a gift." Caelum grinned slightly, a human act he had adapted that involved pulling back his lips somewhat and showing off his front teeth. He purred. Diana pulled the glittering necklaces out of her pocket and clipped the longer one around the dragon's neck. He nuzzled her in thanks as Diana put her own necklace on. "You're welcome," she said as she mounted him. "And thank you, too."
Caelum launched himself into the air, and Diana could feel the air shifting violently around her as the black dragon's great wings flapped. Diana longed to take the goggles off to feel the wind on her face, but then it would just get in her eyes and annoy her. She would have liked to take her helmet off as well and let her hair stream out behind her, but then it would just get tangled. And she would get in big trouble if she were caught. Instead, she left her gear on and tried to think of something to make their trip more exciting.
Suddenly, as if reading her mind, Caelum performed a barrel roll. Diana, completely unprepared, was saved only by her seatbelt-harness that came on all saddles. She laughed. The dragon was known for pulling off sudden aerial stunts, so she was used to it. For the most part. Diana took some deep breaths to calm her pounding heart and the flow of adrenaline running to it.
Looking around her, she saw a familiar dark golden dragon in the distance. Diana nudged Caelum's right side and said, "That way." The dragon understood and swerved in the direction she had gestured towards. He saw the other dragon and recognized her, but Diana said her name, Aura, anyway. Happily, Caelum lifted himself higher into the air and swooped down to Aura's side. He growled to her a greeting, and she purred back.
"Hello Kira," Diana yelled to the rider—it was hard to hear with the helmets and dragons' flapping wings. "Hey, Aura," she added when the golden dragon made a shunning gesture. Aura glanced at Kira and grinned.
Diana had known Kira since fourth grade, and their friendship was as strong as any friendship could be. Kira was not quite two months older than Diana, except on leap years. Diana's birthday was April 29, and Kira's birthday was February 29. When it was not a leap year, Kira celebrated her birthday on the first of March. She liked March more than February. "Fewer hearts and more birds," she had explained. Kira had a great love for birds and knew the scientific names of almost every bird in the region.
Kira looked over at Diana and called back as the wind flicked and fingered her loose black hair, which with her white goggle straps and brown riding suit reminded Diana of a chickadee. Kira pulled her cell phone out and shouted, "Call me!" Then she pointed at a building below, crying a farewell, and signaling to Aura to turn and head down. Diana called a farewell in return and set back in the direction of her home.
In a few minutes, they landed. Diana dismounted and allowed Caelum free roam. If she didn't call beforehand, he'd return at night to sleep.
She slipped inside the house and called out, "I'm home!"
"Hello, Diana," her mother answered from the living room. "Come here, they're showing a report on a new creature they discovered."
Ever since the first dragons had been found, scientists and even ordinary people had been finding more and more new species of a currently known animal or even new species entirely. There were usually periods of half a year to a year between each find. Diana took off her gear, put it in the closet by the door, and dashed to the living room.
Her mother rewound the television clip to the beginning of the segment. A short scene showed footage of the newest species. It looked to Diana to be a cross between a panther and a wolf with a whiplash tail, five or six fine neck spines and long canines like those of a sabre-tooth tiger. With a quiet gasp Diana watched and, as much as she loved them, she hoped they didn't try to sell them as pets. The reporter said it was simply called a 'Sabre-Tooth Dogcat'. The temporary Latin name was felupus ingentidontis, the former word being a mix of 'felis' and 'lupis', Diana assumed, and the latter meaning 'large-tooth'. She figured that scientists would eventually find a better name for the creature though she wanted them to keep 'ingentidontis'. She loved it. It had a nice ring to it.
"That is awesome," she remarked, shifting on the couch. "Thanks for showing me. Now, I met Kira on our flight and told her I'd call her, so I'm going to do that."
"Okay," her mother replied. "When you're done I'd like to see what you got at the Saturday Market!"
"Alright." Diana made for the stairs and ran up to her room, grabbing the portable phone on the way. She opened the door to her bedroom, walked in, closed the door, and flopped down onto her bed. She dialed Kira's cell phone number and waited for her to answer.
The phone stopped ringing and was replaced by a single "Hello?" from Kira.
"Hey, there, it's me," Diana greeted.
"Oh. Hey, Me. How're you doin'?" Kira asked sarcastically. Even after living here in Oregon for so long, some of her old Texan accent showed through when she spoke. Kira had moved up from Texas about six years ago when her parents had decided they were tired of worrying about the thunderstorms and tornadoes. Kira had been downtrodden for months afterward, and she still missed the storms and risks. In her opinion, Oregon was too safe. "I was hoping you'd call me a bit later, I'm shopping right now, but… eh, what the heck."
Diana smiled. "Well. Not only was I wondering if we could get together, but I also bear some interesting news."
"Ooh. Do tell."
Diana began to tell Kira about the new creature that had been discovered, and had great fun letting the words felupus ingentidontis spill off her tongue until Kira asked her to stop. When Diana had finished, Kira replied by simply saying "awesome" and then commencing to tell a story of her own. Apparently some dragons were catching crows and their young for prey, and the crows were learning to recognize dragons as their enemies. So, on their way home one day, Kira and (mainly) Aura had been mobbed by a murder of crows. Kira, who loved crows, was beginning to lose that love. But, she figured, she wouldn't expect any other behavior from them. In short, it was a warning for Diana to relay to Caelum.
Eventually the two began to talk about getting together to hang out, but Kira sadly admitted that she was already booked for the rest of the day. She was going to catch a movie with her family in an hour and then would be going to dinner.
The two friends said goodbye and hung up. Diana frowned and fingered her necklace. She would have to find something else to do. Maybe she and Caelum could go up the woods together? Just up the street an entrance to a forest in the heart of suburbia stood tall and dark, the drooping firs and cedars giving the place an ominous feel. Urban myth claimed the woods were cursed or haunted. Whenever anyone told the legend to Diana she dismissed it with a casual "Pshaw." She had been in there plenty of times, and with everyone else as afraid of it as they were she pretty much had the whole place to herself.
The forest covered quite a large area and could take no less than half an hour to cross completely. Toward the middle, Diana had discovered a fort, a stream with hazelnut trees bordering its boundaries on one side, a fence on the other. The entrance to the fort was marked by two wooden fence-gates flanking a thick board that crossed the stream. Where the gates had come from, Diana had no idea. Upon finding the fort almost exactly two years ago, in the middle of the summer, she fell in love with it and still loved it just as much. On many occasions she'd taken a light folding chair and a book, sitting among the trees and reading.
Diana glanced at the clock—four twenty-three. There was plenty of time before dinner. She asked her mother if a trip to the forest was okay, calling Caelum when they approved before going to change into better woods-trekking clothes. Slipping on her tennis shoes, she went out the door and waited for the black dragon to arrive. After several minutes he still hadn't arrived, so she went to search for him. She found him in his den in the garage, gazing at her with an amused expression. Diana smiled. "Come on, we're going to the woods." Caelum flared his ears, hauled himself up, and walked alongside her partway to the forest. Eventually he got tired of walking and took off into the air, meeting her at the entrance trail. Together, the companions strode into the familiar forest hoping, as they usually did, to find something unfamiliar. Usually their discovery was a new secluded segment of the woodland, a new nest of some sort of bird or dragon, or new climbing trees.
If they had only been able to guess what kind of thing they would discover today.
« Prev Ch 2 of 2

redpajama_set_name: RedPajamaCommonCrawl


text:
Vermintide 2 XBox One launch date & open beta news
Written by Suzie Ford on June 21, 2018 | OpenCritic
Games discussed: Warhammer: Vermintide 2
Fatshark has announced that the XBox One version of its critically-acclaimed Warhammer: Vermintide 2 is going to be released on July 11, 2018. If you can't wait until then, you can check it out as part of the XBox Insider Program starting today during open beta testing.
Preorders for the XBox One version of Vermintide 2 have started as well. The Premium edition yields a number of in-game and real-world perks including a Sigmar Keep Statue and 6x Exclusive Heroic Deeds. You can check out all of the details over at the Microsoft Store.
Vermintide 2 is a visually stunning and groundbreaking melee action game pushing the boundaries of the first person co-op genre. Set during the apocalyptic End Times of the Warhammer world – the game introduces the dark, bloody and twisted forces of Chaos as an enemy faction. Other new features include 15 hero careers, new environments, talent trees and the proprietary Heroic Deeds System.
Suzie Ford


text:
\section{Introduction}

The measure contraction property (MCP) is one of the generalizations of Ricci curvature bounds to non-smooth metric measure spaces $(X,d,\mu)$.
The condition $\mathrm{MCP}(K,N)$ we discuss here was introduced by Ohta \cite{Ohta-MCP} and controls how the measure $\mu$ of a set in $X$ is distorted along geodesics defined by the distance $d$, where $(X,d)$ is assumed to be a length space (see also Sturm \cite{S-II}, for a slightly stronger version of  the same condition).

When $X$ is a complete Riemannian manifold of dimension $N\geq 2$, $d$ is the Riemannian distance, and $\mu$ is the Riemannian measure, the condition $\mathrm{MCP}(K,N)$ is equivalent to require that $X$ has Ricci curvature bounded from below by $K$ (see \cite[Thm. 3.2]{Ohta-MCP}). The measure contraction property is thus a synthetic replacement for Ricci curvature bounds on metric spaces. We refer to Section \ref{s:mms} for a precise definition. We recall that the $\mathrm{MCP}$ is the weakest among the synthetic curvature conditions introduced in \cite{S-I,S-II,LV-Ricci} and developed in the subsequent literature. We stress that all these conditions are stable under pointed Gromov-Hausdorff limits.

\medskip
In this paper, we continue the investigation of measure contraction properties in sub-Riemannian geometry, started in \cite{Juillet,Rifford,MCPc1}.  More precisely we focus our attention on a class of Carnot groups that we call \emph{generalized H-type groups}, which includes the classical H-type Carnot groups, in the sense of Kaplan~\cite{Kaplan,BLU-Stratified}. Here a Carnot group $G$ is considered as a metric measure space $(G,d,\mu)$ equipped with the Carnot-Carath\'eodory distance $d$ and a left-invariant measure $\mu$ (thus a multiple of Haar and Popp measures, see \cite{montgomerybook,BR-popp}).

Even though Carnot Groups (and more in general sub-Riemannian manifolds) can be seen as Gromov-Hausdorff limits of sequences of Riemannian ones with the same dimension, these sequences have Ricci curvature unbounded from below (for an explicit example, see \cite{Rifford}). Thus, it is not clear whether these structures satisfy some $\mathrm{MCP}$.

The simplest Carnot groups, that is the $2d+1$ dimensional Heisenberg groups $\mathbb{H}_{2d+1}$, do not satisfy the $\mathrm{MCP}(K,2d+1)$ for any $K$. Indeed, in \cite{Juillet}, Juillet proved that $\mathbb{H}_{2d+1}$ satisfies
$\mathrm{MCP}(K,N)$ if and only if $K \leq 0$ and $N \geq 2d+3$. The number $2d+3$ is then the optimal dimension for the synthetic condition $\mathrm{MCP}(0,N)$, and is larger than both the topological ($2d+1$) and the Hausdorff ($2d+2$) dimensions of $\mathbb{H}_{2d+1}$.

Juillet's results have been extended in \cite{MCPc1}, where it is proved that any corank $1$ Carnot group $(G,d,\mu)$ of rank $k$ satisfies the $\mathrm{MCP}(K,N)$ if and only if $K \leq 0$ and $N \geq k+3$. In particular, they satisfy the $\mathrm{MCP}(0,k+3)$.

For what concerns more general structures, in \cite{Rifford}, Rifford proved that if a Carnot group of rank $k$ and dimension $n$ is ideal (i.e.\ does not admit abnormal minimizing curves), then it satisfies $\mathrm{MCP}(0,N)$ for some finite $N$ satisfying the inequality $N\geq Q+n-k$, where $Q$ is the Hausdorff dimension of the Carnot group. For a non-ideal Carnot group, if one \emph{assumes} that $\mathrm{MCP}(0,N)$ is satisfied for some $N$, the inequality $N\geq  Q+n-k$ still holds.

\subsection{Geodesic dimension}
Since $\mathrm{MCP}(K,N)$ implies $\mathrm{MCP}(K,N')$ for all $N'\geq N$, the above results raised the question of which is the optimal $N$ such that the $\mathrm{MCP}(K,N)$ is satisfied on a general metric measure space. A bound is given by the concept of geodesic dimension, introduced in \cite{ABR-Curvature} for sub-Riemannian structures and in \cite{MCPc1} for metric measure spaces, and which we briefly overview here (see Section \ref{s:mms} for details). Let $(X,d,\mu)$ be a metric measure space, which for simplicity we assume having negligible cut loci. Let $\Omega \subset X$ be a bounded, measurable set, with $0<\mu(\Omega)<+\infty$, and consider its homothety $\Omega_t$ with center $x$ and ratio $t \in [0,1]$, such that $\Omega_{1}=\Omega$ and $\Omega_{0}=\{x\}$. The \emph{geodesic dimension} $\mathcal{N}(x)$ at the point $x \in X$ is, roughly speaking, the largest integer such that $\mu(\Omega_t) = O(t^{\mathcal{N}(x)})$ when $t \to 0$, for all such $\Omega$.

\begin{prop}[\cite{MCPc1}]\label{t:bound}
Let $(X,d,\mu)$ be a metric measure space with negligible cut loci that satisfies the $\mathrm{MCP}(K,N)$, for some $K \in \mathbb{R}$ and $N >1$ or $K\leq 0$ and $N=1$. Then
\begin{equation}
N \geq \sup \{\mathcal{N}(x) \mid x \in X\}.
\end{equation}
where $\mathcal{N}(x)$ is the geodesic dimension of $X$ at $x$.
\end{prop}

\subsection{Carnot groups}
In the case of left-invariant structures on Carnot (or, more generally, Lie) groups, the number $\mathcal{N}(x)$ is independent on $x\in X$, and we denote it by $\mathcal{N}$.
Furthermore, notice that spaces verifying $\mathrm{MCP}(K,N)$ with $K>0$ are compact, so that Carnot groups can satisfy the $\mathrm{MCP}(K,N)$ only if $K \leq 0$. On the other hand, thanks to the homogeneity properties of Carnot groups, if $\mathrm{MCP}(K,N)$ is satisfied for some $K\leq 0$, then $\mathrm{MCP}(\varepsilon K,N)$ is also satisfied, for all $\varepsilon \geq 0$. For this reason the $\mathrm{MCP}(0,N)$ plays a special role for Carnot groups, which motivated the definition of \emph{curvature exponent}, as
\[
N_0:= \inf\{N>1 \mid \mathrm{MCP}(0,N) \text{ is satisfied}\}.
\]
Indeed, by Proposition~\ref{t:bound}, $N_0 \geq \sup\{ \mathcal{N}(x) \mid x \in X\}$. The following example shows that, in general, the curvature exponent $N_{0}$ can be strictly bigger than the geodesic dimension.
\begin{example}[Riemannian Heisenberg]\label{ex:ex1}
Consider the left-invariant Riemannian structure on $\mathbb{R}^{3}$ generated by the following global orthonormal vector fields:
\begin{equation}
X = \partial_x -\frac{y}{2}\partial_z, \qquad Y = \partial_y + \frac{x}{2}\partial_z, \qquad Z = \partial_z.
\end{equation}
As for any Riemannian structure, $k=n=\mathcal{N} = 3$. In \cite{Rifford} it is proved that, when equipped with the Riemannian volume, this structure satisfies the $\mathrm{MCP}(0,5)$. The same techniques proves that the $\mathrm{MCP}(0,5-\varepsilon)$ is violated for any $\varepsilon>0$, so its curvature exponent is $N_0 = 5$. We stress that this is a left-invariant structure, but is not a Carnot group.
\end{example}

We expect, however, that for Carnot groups the geodesic dimension coincides with the curvature exponent (cf.\ the conjecture formulated in \cite{MCPc1}). Our main result is a proof of this fact for generalized H-type groups (see Section~\ref{s:prel}), which generalizes the results of  \cite{Juillet,MCPc1}, and constitutes then the largest class of structures for which the aforementioned conjecture is verified.

\begin{theorem}\label{t:htype}
Let $(G,d,\mu)$ be a generalized H-type Carnot group of rank $k$ and dimension $n$. Then it satisfies the $\mathrm{MCP}(K,N)$ if and only if $K \leq 0$ and $N \geq k+3(n-k)$, the latter integer being the geodesic dimension of the Carnot group.
\end{theorem}

We observe that generalized H-type groups may have non-trivial abnormal minimizing curves (albeit not strictly abnormal ones). Equivalently, not all generalized H-type groups are ideal, or fat (see Definitions \ref{d:ideal} and \ref{d:fat}).

As a byproduct of the proof, we describe explicitly normal and abnormal geodesics of generalized H-type groups (Lemmas~\ref{l:geods}, \ref{l:abnormals}), the injectivity domain of the exponential map (Lemma \ref{l:ingjd}), and an explicit formula for its Jacobian determinant (Lemma~\ref{l:jacobian}). In particular, this yields the optimal synthesis of generalized H-type groups.

We mention that in \cite{BR-grande1} the Jacobian estimates of this paper have been used to prove a sharp geodesic Brunn-Minkowski inequality for generalized H-type groups, which as a particular case recovers the measure contraction properties of Theorem~\ref{t:htype}.

\subsection{Consequences for optimal transport}

 Theorem~\ref{t:htype} has consequences on the optimal transportation problem on generalized H-type groups, which we briefly outline here.

The well posedness of the Monge problem with cost given by the squared sub-Rieman\-nian distance for classical H-type groups was proved in \cite{Rigot2}, generalizing the previous results for the Heisenberg group obtained in \cite{Rigot1}. The same result for general sub-Riemannian structures whose squared distance is Lipschitz in charts on $X\times X$ was obtained in \cite{AL-transport}. The Lipschitzness assumption on the diagonal was finally removed in \cite{FR}. We refer to \cite{rifford2014sub} for background on the transportation problem in sub-Riemannian geometry.

For all step $2$ sub-Riemannian structures on a smooth manifold $X$ (such as generalized H-type Carnot groups), $d^2$ is locally Lipschitz on $X\times X$. Hence, according e.g.\ to \cite[Theorem 3.2]{FR}, for any pair of compactly supported probability measures $\mu_0,\mu_1 \in \mathcal{P}_{\text{comp}}(X)$, with $\mu_0$ absolutely continuous with respect to a reference measure $\mu$, there exists a unique optimal transport map for the Monge problem with quadratic cost.

In this setting, an interesting problem is to determine whether the Wasserstein geodesic $\mu_t \in \mathcal{P}(X)$, interpolating between $\mu_0$ and $\mu_1$, is absolutely continuous w.r.t.\ $\mu$. An affirmative answer for the $2d+1$ dimensional Heisenberg group was the main result of \cite{FJ}. More general structures have been discussed in \cite[Theorem 3.5]{FR}. There it is proved that, if the sub-Riemannian squared distance is also locally semiconcave outside of the diagonal, then $\mu_t \ll \mu$ for all $t \in [0,1)$. Unfortunately, in presence of non-trivial abnormal minimizers, $d^2$ is not necessarily locally semiconcave outside of the diagonal. Absolute continuity of $\mu_t$ must then be proved directly (see, e.g.\ \cite{BKS17} for the case of corank 1 Carnot groups).

As a consequence of Theorem~\ref{t:htype}, and the fact that step $2$ Carnot groups do not admit branching minimizing geodesics, we can apply \cite[Theorem 1.1]{CM}\footnote{See also the appendix of \cite{CM}, where it is discussed how, under the non-branching assumption, Ohta's $\mathrm{MCP}$, which we use here, is equivalent to Sturm's one, which is the one primarily used in \cite{CM}.}. In particular, we extend the absolute continuity result of \cite{FJ} to a setting with non-trivial abnormal minimizers and higher corank. We refer to \cite{CM} for definitions.
\begin{cor}[\cite{CM}]
Let $(X,d,\mu)$ be a generalized H-type Carnot group. Let $\mu_0,\mu_1 \in \mathcal{P}_2(X)$ with $\mu_0 \ll \mu$. Then there exists a unique $\nu \in \mathrm{OptGeo}(\mu_0,\mu_1)$; it is induced by a map $G: X \to \mathrm{Geo}(X)$, and the unique Wasserstein geodesic  between $\mu_0$ and $\mu_1$ satisfies $(e_t)_\sharp \nu =\rho_t \mu \ll \mu$ for all $t \in [0,1)$. Finally, if $\mu_0,\mu_1$ have bounded support and $\rho_0$ is $\mu$-essentially bounded, then
\begin{equation}
\|\rho_t\|_{L^\infty(X,\mu)} \leq \frac{1}{(1-t)^{k+3(n-k)}} \|\rho_0\|_{L^{\infty}(X,\mu)}, \qquad \forall t \in [0,1).
\end{equation}
\end{cor}

\subsection*{Acknowledgements}
\thanks{This research has been supported by the Grant ANR-15-CE40-0018 of the ANR. This research benefited from the support of the ``FMJH Program Gaspard Monge in optimization and operation research'' and from the support to this program from EDF. This work has been partially supported by the ANR project ANR-15-IDEX-02.}

\bibliographystyle{abbrv}

\section{Preliminaries}\label{s:prel}
We recall some basic facts in sub-Riemannian geometry. For details, see \cite{nostrolibro,rifford2014sub,montgomerybook}.

\subsection{Basic definitions}
A sub-Rieman\-nian manifold is a triple $(M,\mathcal{D},g)$, where $M$ is a smooth, connected manifold with $\dim M=n \geq 3$, $\mathcal{D}$ is a vector distribution of constant rank $k \leq n$ and $g$ is a smooth metric on $\mathcal{D}$. We  assume that the distribution is bracket-generating. A \emph{horizontal curve} $\gamma : [0,1] \to M$ is a Lipschitz continuous path such that $\dot\gamma(t) \in \mathcal{D}_{\gamma(t)}$ for almost every $t$, and we define its \emph{length} as follows
\begin{equation}
\ell(\gamma) = \int_0^1 \sqrt{g(\dot\gamma(t),\dot\gamma(t))}dt.
\end{equation}
The \emph{sub-Rieman\-nian (or Carnot-Carath\'eodory) distance} is defined by:
\begin{equation}\label{eq:infimo}
d(x,y) = \inf\{\ell(\gamma)\mid \gamma(0) = x,\, \gamma(1) = y,\, \gamma \text{ horizontal} \}.
\end{equation}
By Chow-Rashevskii theorem, the bracket-generating condition implies that $d: M \times M \to \mathbb{R}$ is finite and continuous. If the metric space $(M,d)$ is complete, then for any $x,y \in M$ the infimum in \eqref{eq:infimo} is attained. In place of the length $\ell$, we consider the \emph{energy functional}
\begin{equation}
J(\gamma) = \frac{1}{2}\int_0^1 g(\dot\gamma(t),\dot\gamma(t)) dt.
\end{equation}
On the space of horizontal curves defined on a fixed interval and with fixed endpoints, the minimizers of $J$ coincide with the minimizers of $\ell$ parametrized with constant speed. Since $\ell$ is invariant by reparametrization (and we can always reparametrize horizontal curves in such a way that they have constant speed), we do not loose generality in defining \emph{geodesics} as horizontal curves that locally minimize the energy between their endpoints.

The \emph{Hamiltonian} of the sub-Riemannian structure $H : T^*M \to \mathbb{R}$ is defined by
\begin{equation}
H(\lambda) = \frac{1}{2}\sum_{i=1}^k \langle \lambda,X_i\rangle^{2}, \qquad \lambda \in T^*M,
\end{equation}
where $X_1,\dots,X_k$ is any local orthonormal frame for $\mathcal{D}$. Here $\langle \lambda,\cdot\rangle$ denotes the dual action of covectors on vectors. The Hamiltonian vector field $\vec{H}$ is the unique vector field such that $\sigma(\cdot,\vec{H}) = dH$, where $\sigma$ is the canonical symplectic form of the cotangent bundle $\pi:T^*M \to M$. In particular, the \emph{Hamilton equations} are
\begin{equation}\label{eq:Ham}
\dot\lambda(t) = \vec{H}(\lambda(t)), \qquad \lambda(t) \in T^*M.
\end{equation}
If $(M,d)$ is complete, any solution of \eqref{eq:Ham} can be extended to a smooth curve for all times.

\subsection{End-point map and Lagrange multipliers}

Let $\gamma_u :[0,1] \to M$ be an horizontal curve joining $x$ and $y$. Let $X_1,\dots,X_k$ be a smooth orthonormal frame of horizontal vectors fields, defined in a neighborhood of $\gamma_u$, and let $u \in L^\infty([0,1],\mathbb{R}^k)$ be a \emph{control} such that
\begin{equation}
\dot\gamma_u(t) =  \sum_{i=1}^k u_i(t) X_i(\gamma_u(t)), \qquad \text{a.e. } t \in [0,1].
\end{equation}
Let $\mathcal{U} \subset L^\infty([0,1],\mathbb{R}^k)$ be the neighborhood of $u$ such that, for $v \in \mathcal{U}$, the solution of
\begin{equation}
\dot\gamma_v(t) = \sum_{i=1}^k v_i(t) X_i(\gamma_v(t)), \qquad \gamma_v(0) = x,
\end{equation}
is well defined for a.e. $t \in [0,1]$. We define the \emph{end-point map} with base point $x$ as the map $E_{x}: \mathcal{U} \to M$ that sends $v$ to $\gamma_v(1)$. The end-point map is smooth on $\mathcal{U}$.

We can consider $J : \mathcal{U} \to \mathbb{R}$ as a smooth functional on $\mathcal{U}$ (identifying $\mathcal{U}$ with a neighborhood of $\gamma_u$ in the space of horizontal curves starting from $x$). A minimizing geodesic $\gamma_u$ is a solution of the constrained minimum problem
\begin{equation}
\min\{J(v) \mid  v \in \mathcal{U},\, E_x(v) = y\}.
\end{equation}
By the Lagrange multipliers rule, there exists a non-trivial pair $(\lambda_1,\nu)$, such that
\begin{equation}\label{eq:multipliers}
\lambda_1 \circ D_u E_x  = \nu D_u J, \qquad \lambda_1 \in T_y^*M, \qquad\nu \in \{0,1\},
\end{equation}
where $\circ$ denotes the composition of linear maps and $D$ the (Fr\'echet) differential. If $\gamma_u : [0,1] \to  M$ with control $u \in \mathcal{U}$ is an horizontal curve (not necessarily minimizing), we say that a non-zero pair $(\lambda_1,\nu) \in T_y^*M \times \{0,1\}$ is a \emph{Lagrange multiplier} for $\gamma_u$ if \eqref{eq:multipliers} is satisfied. The multiplier $(\lambda_1,\nu)$ and the associated curve $\gamma_u$ are called \emph{normal} if $\nu = 1$ and \emph{abnormal} if $\nu = 0$. Observe that Lagrange multipliers are not unique, and a horizontal curve may be both normal \emph{and} abnormal. Observe also that $\gamma_u$ is an abnormal curve if and only if $u$ is a critical point for $E_x$. In this case, $\gamma_u$ is also called a \emph{singular curve}. The following characterization is a consequence of the Pontryagin Maximum Principle.
\begin{theorem} \label{t:utile}
Let $\gamma_u :[0,1] \to M$ be an horizontal curve joining $x$ and $y$. A non-zero pair $(\lambda_1,\nu) \in T_y^*M \times \{0,1\}$ is a Lagrange multiplier for $\gamma_u$ if and only if there exists a Lipschitz curve $\lambda(t) \in T_{\gamma_u(t)}^*M$ with $\lambda(1) = \lambda_1,$ such that
\begin{itemize}
\item if $\nu = 1$ then $\dot{\lambda}(t) = \vec{H}(\lambda(t))$, i.e.\ it is a solution of Hamilton equations,
\item if $\nu =0 $ then $\sigma(\dot\lambda(t), T_{\lambda(t)} \mathcal{D}^\perp) = 0$,
\end{itemize}
where $\mathcal{D}^\perp \subset T^*M$ is the sub-bundle of covectors that annihilate the distribution.
\end{theorem}
In the first (resp.\ second) case, $\lambda(t)$ is called a \emph{normal} (resp.\ \emph{abnormal}) \emph{extremal}. Normal extremals are   integral curves $\lambda(t)$ of $\vec{H}$. As such, they are smooth, and characterized by their \emph{initial covector} $\lambda = \lambda(0)$. A geodesic is normal (resp.\ abnormal) if admits a normal (resp.\ abnormal) extremal. On the other hand, it is well known that the projection $\gamma_\lambda(t) = \pi(\lambda(t))$ of a normal extremal is locally minimizing, hence it is a normal geodesic. The \emph{exponential map} at $x \in M$ is the map
\begin{equation}
\exp_x : T_x^*M \to M,
\end{equation}
which assigns to $\lambda \in T_x^*M$ the final point $\pi(\lambda(1))$ of the corresponding normal geodesic. The curve $\gamma_\lambda(t):=\exp_x(t \lambda)$, for $t \in [0,1]$, is the normal geodesic corresponding to $\lambda$, which has constant speed $\|\dot\gamma_\lambda(t)\| = \sqrt{2H(\lambda)}$ and length $\ell(\gamma|_{[t_1,t_2]}) = \sqrt{2H(\lambda)}(t_2-t_1)$.

\begin{definition}\label{d:ideal}
A sub-Riemannian structure $(M,\mathcal{D},g)$ is \emph{ideal} if the metric space $(M,d)$ is complete and there exists no non-trivial abnormal minimizers.
\end{definition}
\begin{definition}\label{d:fat}
A sub-Riemannian structure $(M,\mathcal{D},g)$ is \emph{fat} if for all $x \in M$ and $X \in \mathcal{D}$, $X(x) \neq0$, then $\mathcal{D}_x + [X,\mathcal{D}]_x=T_xM$.
\end{definition}
The definition of ideal structures appears in \cite{Rifford,rifford2014sub}, in the equivalent language of singular curves. We stress that fat sub-Riemannian structures admit no non-trivial abnormal curves (see \cite[Section 5.6]{montgomerybook}). In particular, complete fat structures are ideal.

\section{Generalized H-type Carnot groups}
A \emph{Carnot group $(G,\star)$ of step $s$} is a connected, simply connected Lie group of dimension $n$, such that its Lie algebra $\mathfrak{g} = T_e G$ is stratified and nilpotent of step $s$, that is
\begin{equation}
\mathfrak{g} = \mathfrak{g}_1 \oplus \dots \oplus \mathfrak{g}_s,
\end{equation}
with
\begin{equation}
[\mathfrak{g}_1,\mathfrak{g}_j] = \mathfrak{g}_{1+j},\quad \forall 1\leq j\leq s, \quad \mathfrak{g}_s \neq \{0\}, \quad \mathfrak{g}_{s+1} =\{0\}.
\end{equation}

Let $\mathcal{D}$ be the left-invariant distribution generated by $\mathfrak{g}_1$, with a left-invariant scalar product $g$.  This defines a sub-Riemannian structure $(G,\mathcal{D},g)$ on the Carnot group. For $x \in G$, we denote with $\tau_x(y) := x \star y$ the left translation. The map $\tau_x : G \to G$ is a smooth isometry.  Any Carnot group, equipped with the Carnot-Carath\'eodory distance $d$ and the Lebesgue measure $\mu$ of $G=\mathbb{R}^n$ is a complete metric measure space $(X,d,\mu)$. Moreover, a Carnot group is fat if and only if it is ideal \cite[Section 8.2]{MCPc1}.

\subsection{Generalized H-type groups}

Let $(G,\star)$ be a step $2$ Carnot group, with Lie algebra $\mathfrak{g}$ of rank $k$ and dimension $n$. In particular $\dim\mathfrak{g}_1=k$, $\dim\mathfrak{g}_2=n-k$, and
\begin{equation}
[\mathfrak{g}_1,\mathfrak{g}_1] = \mathfrak{g}_2, \qquad [\mathfrak{g}_i,\mathfrak{g}_2]=0, \qquad i= 1,2.
\end{equation}
Fix a scalar product $g$ on $\mathfrak{g}$ such that $\mathfrak{g}_1 \perp \mathfrak{g}_2$. The restriction $g|_{\mathfrak{g}_1}$ induces a left-invariant sub-Riemannian structure $(\mathcal{D},g)$ on $G$, such that $\mathcal{D}(p) = \mathfrak{g}_1(p)$ for all $p \in G$. For any $V \in  \mathfrak{g}_2$, the skew-symmetric operator $J_V: \mathfrak{g}_1 \to \mathfrak{g}_1$ is defined by
\begin{equation}
g(X,J_V Y) = g(V,[X,Y]), \qquad \forall X,Y \in \mathfrak{g}_1.
\end{equation}
With an abuse of language we still call $(G,\star)$, equipped with a scalar product $g$ as above and the induced left-invariant sub-Riemannian structure, a step $2$ Carnot group.
\begin{definition} \label{d:genht}
A step $2$ Carnot group is of \emph{generalized H-type} if there exists a symmetric, non-zero and non-negative operator $S : \mathfrak{g}_1 \to \mathfrak{g}_1$ such that
\begin{equation}\label{eq:13}
J_V J_W + J_W J_V = -2 g(V,W) S^2, \qquad \forall V,W \in \mathfrak{g}_2.
\end{equation}
\end{definition}
\begin{rmk}
By polarization, it is easy to show that a step $2$ Carnot group is of generalized H-type if and only if there exists a symmetric, non-negative and non-zero operator $S:\mathfrak{g}_1 \to \mathfrak{g}_1$  such that $J_V^2 = - \|V\|^2 S^2$ for all $V \in \mathfrak{g}_2$.
\end{rmk}
\begin{rmk}
The case $S = \mathrm{Id}_{\mathfrak{g}_1}$ corresponds to classical H-type groups, introduced by Kaplan in \cite{Kaplan}. We refer to \cite[Chapter 18]{BLU-Stratified} for a modern introduction.
\end{rmk}
\begin{rmk}
It is easy to check that corank $1$ Carnot groups, that is step $2$ Carnot groups with $n=k+1$, are always of generalized H-type. Furthermore, if the operator $S$ is non-degenerate (and thus $k=2d$ is even and $S>0$), we are in the case of contact Carnot groups. The eigenvalues of $S$ are also called \emph{frequencies} of the Carnot group (see \cite{ABB-Hausdorff,MCPc1}).
\end{rmk}

\subsection{On existence}
As discussed in \cite[Chapter 18]{BLU-Stratified} and proved in \cite[Corollary 1]{Kaplan}, as a consequence of the Clifford algebra structure enforced by \eqref{eq:13}, classical H-type Carnot groups of rank $k$ and dimension $n$ exist if and only if $n-k \leq \rho(k)$, where $\rho$ is the Hurwitz-Radon function\footnote{$\rho(N)-1$ gives the maximal number of independent vector fields on $\mathbb{S}^{N-1}$.}. In particular, H-type groups of odd rank do not exist (indeed, when $S$ is the identity, $J_V$ must be at the same time skew-symmetric and orthogonal).

The class of generalized H-type groups is larger, and now discuss their existence. Let $0<\lambda_1<\dots<\lambda_\ell$ be the non-zero eigenvalues of $S$, and let $E_1,\dots,E_\ell$ be the corresponding eigenspaces. For all $V \in \mathfrak{g}_2$, we have $J_V^2 = - \|V\|^2 S^2$. Thus $J_V$ commutes with $S^2$, hence with $S$. We can then split the skew-symmetric operator $J_V$ in the orthogonal direct sum
\begin{equation}
J_V = \bigoplus_{i=1}^\ell \lambda_i J_V^i, \qquad J_V^i : = \frac{1}{\lambda_i} J_V|_{E_i}.
\end{equation}
The generalized H-type condition \eqref{eq:13} is equivalent to the fact that each linear map $\mathfrak{g}_2 \ni V \mapsto J_V^i$ must satisfy a standard H-type condition in dimension $k_i = \dim E_i$. By the classical results mentioned above, this is possible if and only if for all $i=1,\dots,\ell$, the condition $n-k < \rho(k_i)$ is satisfied. Hence there exists a generalized H-type Carnot group of dimension $n$, rank $k$, with prescribed eigenvalues $\lambda_i$ and multiplicities $k_i$ if and only if
\begin{equation}
n-k < \min\{\rho(k_i)\mid i=1,\dots,\ell\}.
\end{equation}
Notice that the dimension of the kernel of $S$ does not play any role in this condition. As a consequence, generalized H-type Carnot groups with odd rank do exist.

\subsection{Coordinate presentation} \label{s:normalforms}

The group exponential map $\mathrm{exp}_{G} : \mathfrak{g} \to G$ associates with $V \in \mathfrak{g}$ the element $\gamma_V(1)$, where $\gamma_V: [0,1] \to G$ is the unique integral curve of the vector field defined by $V$ such that $\gamma_V(0) = 0$. Since $G$ is simply connected and $\mathfrak{g}$ is nilpotent, $\mathrm{exp}_G$ is a smooth diffeomorphism.
The choice of an orthonormal basis $e_1,\dots,e_k$ on $\mathfrak{g}_1$ and $f_1,\dots,f_{n-k}$ on $\mathfrak{g}_2$ yields, through the group exponential map, analytic coordinates $(x,z) \in \mathbb{R}^k \times \mathbb{R}^{n-k}$ on $G$ such that $p =(x,z)$ if and only if
\begin{equation}
p = \mathrm{exp}_{G} \left(\sum_{i=1}^{k} x_i e_i + \sum_{\alpha=1}^{n-k} z_\alpha f_\alpha\right).
\end{equation}
Since $[\mathfrak{g}_1,\mathfrak{g}_1] = \mathfrak{g}_2$, there exists a linearly independent family of $k\times k$ skew-symmetric matrices $L^\alpha$, for $\alpha=1,\dots,n-k$, such that the Lie product on $\mathfrak{g}$ reads
\begin{equation}
[e_i,e_j]_{\mathfrak{g}} = \sum_{\alpha=1}^{n-k} L_{ij}^\alpha f_\alpha, \qquad i,j=1,\dots,k.
\end{equation}
In these coordinates, the sub-Riemannian distribution on $G$ is generated by the following set of global orthonormal left-invariant vector fields
\begin{equation}\label{eq:fields}
X_i = \partial_{x_i} - \frac{1}{2}\sum_{\alpha=1}^{n-k}\sum_{j=1}^{k} L_{ij}^\alpha x_j \partial_{z_\alpha}, \qquad i =1,\dots, k,
\end{equation}
obtained by left translation of the corresponding $e_i$. Furthermore, $\partial_{z_\alpha}$ are left-invariant vector fields obtained by left translation of the basis $f_\alpha$ of $\mathfrak{g}_2$.

Observe in particular that
\begin{equation}
[X_i,X_j] = \sum_{\alpha=1}^{n-k}L^{\alpha}_{ij} \partial_{z_\alpha}, \qquad  i,j = 1,\dots,k.
\end{equation}
If $V = \sum v_\alpha f_\alpha$, then $J_V$ is represented by the skew-symmetric matrix $L_v:= \sum v_\alpha L^\alpha$. Notice that $v \mapsto L_v$ is linear. Property \eqref{eq:13} corresponds to the following relations:
\begin{equation}
L_v L_w + L_w L_v = -2 (v \cdot w) S^{2}, \qquad \forall v,w \in \mathbb{R}^{n-k},
\end{equation}
where the dot denotes the Euclidean scalar product. In particular, $L_v$ and $L_w$ anti-commute when $v \perp w$, and $L_v^2= -|v|^2 S^2$. By a suitable orthogonal transformation of the $e_i$'s, we assume that $S$ is diagonal. Since $L_v^2 = - |v|^2 S^2$ and $L_v$ is skew-symmetric the non-zero eigenvalues of $S$ appear in pairs and, since $S \geq 0$, we have
\begin{equation}\label{eq:normalformS}
S = \begin{pmatrix}
\mathbbold{0}_{k-2d} & & & \\
& \alpha_1 \mathbbold{1} & & \\
& & \ddots & \\
& & & \alpha_d \mathbbold{1}
\end{pmatrix}, \qquad \mathbbold{1} = \begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix},
\end{equation}
for some $1\leq d \leq k$, and with $0<\alpha_1\leq \alpha_2\leq \dots\leq \alpha_d$. As a consequence, for all $v \in \mathbb{R}^{n-k}$, the real normal form $\tilde{L}_v$ of $L_v$ is
\begin{equation}\label{eq:normalformlLv}
\tilde{L}_v = \begin{pmatrix}
\mathbbold{0}_{k-2d} & & & \\
& |v|\alpha_1 \mathbbold{J} & & \\
& & \ddots & \\
& & & |v|\alpha_d \mathbbold{J}
\end{pmatrix}, \qquad \mathbbold{J} = \begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix},
\end{equation}
and its spectrum is $\sigma(L_v) = \{0,\pm i |v|\alpha_1,\dots,\pm i |v|\alpha_d \}$, where the zero value is not present if $k=2d$ (that is, when $S>0$).

Henceforth, we fix a choice of exponential coordinates $(x,z)$, in such a way that the sub-Riemannian structure on $G$ is defined by the left-invariant fields \eqref{eq:fields}, and \eqref{eq:normalformS} holds.

\subsection{Lie group isomorphisms vs sub-Riemannian isometries}
Notice that all generalized H-type groups of fixed rank $k$ and dimension $n$ are isomorphic as Lie groups, but not isometric as sub-Riemannian structures.

Let $G$ and $G'$ be generalized H-type groups and $\phi: G\to G'$ be a Lie group isomorphism. In particular $\phi_{*}:\mathfrak{g}\to \mathfrak{g}'$ is a Lie algebra isomorphism preserving the grading and we denote by $M:\mathfrak{g}_{1}\to \mathfrak{g}_{1}'$ and $T  :\mathfrak{g}_{2}\to \mathfrak{g}_{2}'$ its restrictions. If $\phi$ is also a sub-Riemannian isometry (and thus $M:\mathfrak{g}_{1}\to \mathfrak{g}_{1}'$ is an isometry of inner product spaces), then
\begin{equation}
S'=\frac{\Tr(TT^{*})}{n-k}MSM^*,
\end{equation}
where $A^{*}$ denotes the adjoint of a linear operator $A$ with respect to the scalar products on the corresponding spaces.

It follows that the ratios between the non-zero eigenvalues $\alpha_{i}$, with $i=1,\ldots,d$, of the symmetric operator $S$ and the dimensions of the corresponding eigenspaces are preserved by isometries and distinguish between non-isometric sub-Riemannian structures.

This can be seen already in the subclass of contact Carnot groups. In this case the structure of the cut locus (see \cite{BBN16}) and the isometry group of the structure (see \cite{LR17}) highly depend on the ratio $\alpha_{1}/\alpha_{2}$.

\section{Measure contraction property and geodesic dimension}\label{s:mms}

Let $(X,d,\mu)$ be a metric measure space, where $(X,d)$ is a length space and $\mu$ be a Borel measure such that $0< \mu(B(x,r)) < +\infty$ for any metric ball $B(x,r)$.
Moreover let us assume that $(X,d)$ has negligible cut loci: for any $x \in X$ there exists a negligible set $\mathcal{C}(x)$ and a measurable map $\Phi^x : X \setminus \mathcal{C}(x) \times [0,1] \to X$, such that the curve $\gamma(t)=\Phi^x(y,t)$ is the unique minimizing geodesic from $x$ with $y$.

For any set $\Omega$, we consider its geodesic homothety of center $x \in X$ and ratio $t \in [0,1]$:
\begin{equation}\label{eq:omot}
\Omega_t:= \{\Phi^x(y,t) \mid y \in \Omega \setminus \mathcal{C}(x)\}.
\end{equation}
For any $K \in \mathbb{R}$, define the function
\begin{equation}\label{eq:sturm}
s_K(t):= \begin{cases}
(1/\sqrt{K}) \sin(\sqrt{K} t) & \text{if } K>0, \\
t & \text{if } K=0, \\
(1/\sqrt{-K}) \sinh(\sqrt{-K} t) & \text{if } K<0.
\end{cases}
\end{equation}
\begin{definition}[Ohta \cite{Ohta-MCP}]\label{d:MCP}
Let $K \in \mathbb{R}$ and $N>1$, or $K \leq 0$ and $N=1$. We say that $(X,d,\mu)$ satisfies the \emph{measure contraction property} $\mathrm{MCP}(K,N)$ if for any $x \in M$ and any measurable set $\Omega$ with  with $0< \mu(\Omega)< + \infty$ (and with $\Omega \subset B(x,\pi\sqrt{N-1/K})$ if $K > 0$)
\begin{equation}\label{eq:MCP}
\mu(\Omega_{t}) \geq \int_{\Omega} t \left[\frac{s_K(t d(x,z)/\sqrt{N-1})}{s_K(d(x,z)/\sqrt{N-1})}\right]^{N-1} d\mu(z) , \qquad \forall t \in [0,1],
\end{equation}
where we set $0/0 = 1$ and the term in square bracket is $1$ if $K \leq 0$ and $N=1$.
\end{definition}
Ohta defines the $\mathrm{MCP}$ for general length spaces, possibly with non-negligible cut loci. In our setting, Definition~\ref{d:MCP} is equivalent to Ohta's, see \cite[Lemma 2.3]{Ohta-MCP}.

\subsection{The geodesic dimension}

The geodesic dimension was introduced in \cite{ABR-Curvature} for sub-Riemannian structures and extended in \cite{MCPc1} to the more general setting of metric measure spaces (with negligible cut loci).
\begin{definition} \label{d:gd}
Let $(X,d,\mu)$ be a metric measure space with negligible cut locus. For any $x \in X$ and $s > 0$, define
\begin{equation}\label{eq:criticalratio}
C_s(x):=\sup\left\lbrace \limsup_{t \to 0^+} \frac{1}{t^s}\frac{\mu(\Omega_t)}{\mu(\Omega)} \mid \Omega \text{ measurable, bounded, $0<\mu(\Omega)<+\infty$}\right\rbrace,
\end{equation}
where $\Omega_t$ is the homothety of $\Omega$ with center $x$ and ratio $t$ as in \eqref{eq:omot}. The \emph{geodesic dimension} of $(X,d,\mu)$ at $x \in X$ is the non-negative real number
\begin{equation}\label{eq:gddefs}
\mathcal{N}(x)  := \inf \{s > 0 \mid C_s(x) = +\infty \} = \sup \{s > 0 \mid C_s(x) = 0 \},
\end{equation}
with the conventions $\inf \emptyset = + \infty$ and $\sup \emptyset = 0$.
\end{definition}
Roughly speaking, the measure $\mu(\Omega_t)$ vanishes at least as $t^{\mathcal{N}(x)}$ or more rapidly, for $t \to 0$. The two definitions in \eqref{eq:gddefs} are equivalent since $s \geq s'$ implies $C_s(x) \geq C_{s'}(x)$.

When $(X,d,\mu)$ is a metric measure space defined by an equiregular sub-Rieman\-nian or Riemannian structure, equipped with a smooth measure $\mu$, we have
\begin{equation}
\mathcal{N}(x) \geq \dim_H(X) \geq \dim(X), \qquad \forall x \in X,
\end{equation}
and both equalities hold if and only if $(X,d,\mu)$ is Riemannian (see \cite[Proposition 5.49]{ABR-Curvature} and \cite{MCPc1}). Here $\dim_H(X)$ denotes the Hausdorff dimension of the metric space $(X,d)$. We refer to \cite{MCPc1} for a discussion of the other properties of the geodesic dimension. We only remark that, in the case of left-invariant structure on Carnot (or, more generally, Lie) groups, thanks to left-invariance, the number $\mathcal{N}(x)$ is independent of $x\in X$, and we denote it by $\mathcal{N}$. As a consequence of the explicit formula for $\mathcal{N}$ on sub-Riemannian structures \cite[Section 7]{MCPc1}, we have the following.
\begin{lemma}\label{l:fact}
The geodesic dimension of generalized H-type Carnot groups is $k + 3(n-k)$.
\end{lemma}

\begin{rmk}
For simplicity, and since all step $2$ sub-Riemannian structures satisfy this assumption, we defined the geodesic dimension and the measure contraction property for metric measure spaces with negligible cut loci. Nevertheless, these notions can be generalized to any metric measure space with little effort. Finally, let us recall that it is not known whether general Carnot groups have negligible cut loci (this is related with the Sard conjecture in sub-Riemannian geometry \cite{AAAopenproblems,RT-MorseSard,DMOPV-Sard}).
\end{rmk}

\section{Geodesics of generalized H-type Carnot groups} \label{s:five}

In this section we compute the explicit expression of geodesics in generalized H-type Carnot groups. By left-invariance, it is sufficient to consider geodesics starting from the identity $e = (0,0)$. We denote covectors $\lambda = \sum_{i=1}^k u_i dx_i + \sum_{\alpha=1}^{n-k} v_\alpha dz_\alpha \in T^*_eG$, with $(u,v) \in \mathbb{R}^k \times \mathbb{R}^{n-k}$ coordinates on $T^*_e G$ dual to the exponential coordinates $(x,z) \in \mathbb{R}^k \times \mathbb{R}^{n-k}$ of $G$ (see Section~\ref{s:normalforms}).

\addtocontents{toc}{\gobblefive} \subsection*{Convention} For any analytic function $f: \mathbb{C} \to \mathbb{C}$ and any square matrix $M$, the expression $f(M)$ denotes the matrix valued function defined by the Taylor series of $f$ at zero. For example,
\begin{equation}
\frac{\sin M}{M} := \sum_{\ell=0}^{\infty} \frac{M^{2\ell}}{(2\ell+1)!},
\end{equation}
which is well defined for any $M$, possibly degenerate. With this convention, we routinely simplify matrix computations by reducing to functional calculus, and then evaluating on matrices. For example, for any pair $f,g$ of analytic functions, $f(M) g(M) = (fg)(M)$.
Furthermore, for $s \in \mathbb{R}$ we observe that
\begin{equation}\label{eq:derivatio}
\partial_s (f(s M)) = f'(s M) M,
\end{equation}
and similarly for integration. If $M$ can be diagonalized and has eigenvalues $\lambda_i \in \mathbb{C}$, then
\begin{equation}
\det(f(M)) = \prod_i f(\lambda_i).
\end{equation}

\begin{lemma}[Normal geodesics]\label{l:geods}
The exponential map $\exp_e :T^*_e G \to G$ of a generalized H-type Carnot group is given by $\exp_e(u,v) = (x,z)$, where
\begin{align}
x & = f(L_v) u, \\
z & = \left(\frac{g(L_v) u \cdot u }{2|v|}\right) \frac{v}{|v|},
\end{align}
where $|\cdot|$ is the Euclidean norm and $f,g$ are the analytic functions
\begin{equation}\label{eq:fg}
f(z) = \frac{1-e^{-z}}{z}, \qquad g(z) = 1 -\frac{\sinh(z)}{z}.
\end{equation}
Geodesics $\gamma_\lambda:[0,1] \to G$ with initial covector $\lambda =(u,v)$ are given by $\gamma_\lambda(t) = \exp_e(tu,tv)$.
\end{lemma}
\begin{rmk}\label{eq:geodesicsclassical}
For classical H-type groups, $L_v^2 = -|v|^2 \mathbbold{1}$, and the above expressions yield
\begin{align}
x & = \left(\frac{\sin |v|}{|v|}\mathbbold{1} + \frac{\cos |v|-1}{|v|^2}L_{v} \right) u ,\\
z  & = |u|^2\left(\frac{|v|-\sin |v|}{2|v|^2}\right)\frac{v}{|v|}.
\end{align}
If $v= 0$, our convention implies $\exp_e(u,0) = (u,0)$.
\end{rmk}
\begin{proof}
Let $h_x = (h_x^1,\dots,h_x^k) : T^*G \to \mathbb{R}^k$ and $h_z = (h_z^1,\dots,h_{z}^{n-k}) : T^*G \to \mathbb{R}^{n-k}$, where $h_x^{i}(\lambda):= \langle \lambda, X_i\rangle$, for $i=1,\dots,k$ and $h_z^\alpha(\lambda) := \langle \lambda, \partial_{z_\alpha}\rangle$ for all $\alpha =1,\dots,n-k$. Thus, $H = \frac{1}{2}|h_x|^2$. Hamilton equations are
\begin{equation}
\dot{h}_z = 0, \qquad
\dot{h}_x = -L_{h_z} h_x, \qquad \dot{x} = h_x, \qquad \dot{z}^\alpha = -\tfrac{1}{2}h_x \cdot L^\alpha x,
\end{equation}
the dot denoting the derivative with respect to $t$. Set $h_z(0)=v$ and $h_x(0) =u$. Then,
\begin{equation}
h_z(t) = v, \qquad h_x(t) = e^{-t L_v} u.
\end{equation}
The equations for $(x,z)$ can be easily integrated. Exploiting our convention, we obtain
\begin{equation}
x(t) = \int_0^t e^{-s L_v} u \, ds = \frac{\mathbbold{1} - e^{-t L_v}}{L_v} u = f(t L_v) tu.
\end{equation}
To deal with the $z$ coordinate, observe that for any $w\perp v \in \mathbb{R}^{n-k}$, we have
\begin{equation}
2w \cdot \dot{z} = - h_x \cdot L_w x = \sum_{m,n=0}^{\infty} c_{m,n} t^{n+m} (L_v^m u)\cdot (L_w L_v^n u),
\end{equation}
for some coefficients $c_{m,n}$. We claim that each term $(L_v^m u)\cdot (L_w L_v^n u) = 0$, and thus the above expression vanishes identically. To prove this claim, assume without loss of generality that $m\geq n$. The argument is different depending on whether the difference $m-n$ is even or odd. If $m=n+2\ell$, we have
\begin{equation}
(L_v^{n+2\ell} u) \cdot (L_w L_v^n u) = (-1)^\ell (L_v^{n+\ell} u) \cdot (L_v^\ell L_w L_v^n u) = (L_v^{n+\ell}u) \cdot (L_w L_v^{n+\ell}u) = 0.
\end{equation}
If $m = n+2\ell +1$, on the other hand, we have
\begin{multline}
(L_v^{n+2\ell+1}u)\cdot(L_w L_v^n u) = -(L_v^n u) \cdot (L_v^{2\ell+1} L_w L_v^n u) = (L_v^{n} u)\cdot (L_w L_v^{n+2\ell+1} u) = \\
= -(L_w L_v^{n} u) \cdot (L_v^{n+2\ell+1}u) = -(L_v^{n+2\ell+1}u) \cdot (L_w L_v^{n} u),
\end{multline}
which implies $(L_v^{n+2\ell+1}u)\cdot(L_w L_v^n u)= 0$, and proves the claim.

As a consequence of the claim, $z(t)$ is parallel to $v$. Then, we have
\begin{align}
\dot{z}(t) & = \frac{\dot{z}(t)\cdot v}{|v|}\frac{v}{|v|} = -\frac{h_x(t) \cdot L_v x(t) }{2|v|}\frac{v}{|v|} \\
& = -\frac{t L_v f(tL_v)e^{t L_v}u \cdot u}{2|v|} \frac{v}{|v|} \\
& = \frac{[\mathbbold{1}-\cosh(t L_v)]u \cdot u}{2|v|} \frac{v}{|v|},
\end{align}
where in the second line we used that $L_v$ is skew-symmetric, and in the third line we replaced the matrix $tL_v f(tL_v)e^{t L_v}$ with its symmetric part, which can be easily computed from the explicit form of $f$. Integration on the time interval $[0,1]$ concludes the proof.
\end{proof}

Consider a non-zero covector $\lambda = (u,v)$, with $u \in \ker S$. Since $Su=0$ implies $L_v u = 0$, by Lemma~\ref{l:geods}, these curves are straight lines of the form
\begin{equation}
\exp_e(t u,tv) = (tu,0), \qquad \forall v \in \mathbb{R}.
\end{equation}
In particular, there is an infinite number of initial covectors yielding the same geodesic, which is thus abnormal. All abnormal geodesics are of this type.

\begin{lemma}[Abnormal curves and abnormal geodesics]\label{l:abnormals} Abnormal curves of generalized H-type groups are the Lipschitz curves $\gamma_u(t):[0,1] \to G$ with control $u \in L^\infty([0,1],\mathbb{R}^k)$ such that $u(t) \in \ker S$. The curve $\gamma_u(t)$ is an abnormal geodesic if and only if it is a normal geodesic with initial covector $\lambda = (u,v)$ with $u \in \ker S$, i.e.
\begin{equation}
\exp_e(t u,tv) = (tu,0), \qquad \forall v \in \mathbb{R}.
\end{equation}
\end{lemma}
\begin{proof}
We use the explicit characterization of abnormal curves of Theorem~\ref{t:utile}. Let $h_x = (h_x^1,\dots,h_x^k) : T^*G \to \mathbb{R}^k$ and $h_z = (h_z^1,\dots,h_{z}^{n-k}) : T^*G \to \mathbb{R}^{n-k}$, where $h_x^{i}(\lambda):= \langle \lambda, X_i\rangle$, for $i=1,\dots,k$ and $h_z^\alpha(\lambda) := \langle \lambda, \partial_{z_\alpha}\rangle$ for all $\alpha =1,\dots,n-k$. In these coordinates,
\begin{equation}
\mathcal{D}^\perp = \{(x,z,h_x,h_z) \in T^*G  \mid h_x = 0\}.
\end{equation}
Let $\gamma_u(t)$ be an abnormal geodesic, with $u \in L^\infty([0,1],\mathbb{R}^k)$. Let $\lambda(t) = (x(t),z(t),0,h_z(t))$ be a Lipschitz curve in $\mathcal{D}^\perp$ such that $\pi(\lambda(t)) = \gamma_u(t)$. In particular
\begin{equation}
\dot{\lambda}(t) = \sum_{i=1}^k u_i(t) X_i + \sum_{\alpha = 1}^{n-k} \dot{h}_z^\alpha(t) \partial_{h_{z}^\alpha},
\end{equation}
where the vector field $X_i$ in the r.h.s.\ is regarded as a vector field on $T^*G$. We now demand that $\sigma_{\lambda(t)}(\dot\lambda,T_{\lambda}\mathcal{D}^\perp) = 0$. First, since $\partial_{z_\alpha}|_{\lambda(t)} \in T_{\lambda(t)}\mathcal{D}^\perp$, we obtain the condition
\begin{equation}
0 = \sigma_{\lambda(t)}(\dot\lambda,\partial_{z_\alpha}) = \dot{h}_{z}^\alpha(t), \qquad \forall \alpha = 1,\dots,n-k,\quad \text{a.e. } t \in [0,1].
\end{equation}
Hence $h_z(t)$ is constant. This constant must be $h_z(t) = v \neq 0$, otherwise the associated abnormal Lagrange multiplier $\lambda_1 = \lambda(1) \in \mathcal{D}^\perp_{(x,z)} \subset T_{(x,z)}^*G$ would be the zero covector, which is impossible by Theorem~\ref{t:utile}. Furthermore, since $X_j|_{\lambda(t)} \in T_{\lambda(t)}\mathcal{D}^\perp$, we have
\begin{equation}
0 = \sigma_{\lambda(t)}(\dot\lambda,X_j) = \sum_{i=1}^k \sum_{\alpha=1}^{n-k} u_i(t) L_{ij}^\alpha v_\alpha, \qquad \forall j=1,\dots,k,\quad \text{a.e. } t \in [0,1],
\end{equation}
which implies $u(t) \in \ker L_v$. Since $v \neq 0$ and $L_v^2 = -|v|^2 S^2$, this condition is equivalent to $u(t) \in \ker S$. Thus, an horizontal curve $\gamma_u(t)$ is abnormal if and only if it satisfies
\begin{equation}
\dot{\gamma}_u(t) = \sum_{i=1}^k u_i(t) X_i(\gamma_{u}(t)), \qquad u(t) \in \ker S, \qquad u \in L^\infty([0,1],\mathbb{R}^k).
\end{equation}
Using the explicit form of the vector fields $X_i$, this implies $x(t) \in \ker S$ and, in turn, $z(t) = 0$. In particular $\gamma_u(t) = \int_0^t u(s) ds$ lies in the plane $\mathbb{R}^k \subset \mathbb{R}^k \oplus \mathbb{R}^{n-k}$ of $G$, and its length (resp.\ energy) is equal to its Euclidean length (resp.\ energy) in $\mathbb{R}^k$. If we demand $\gamma_u(t)$ to be a geodesic, i.e.\ a locally energy minimizing curve, we have that, $x(t)$ must be a straight line of the form $x(t) = tu$, in particular $u(t) = u \in \ker S$. From Lemma~\ref{l:geods}, we observe that the abnormal geodesics $x(t)=(tu,0)$, with $u \in \ker S$, are also normal, with initial covectors $\lambda = (u,v)$, for any $v \in \mathbb{R}^{n-k}$.
\end{proof}

We conclude this section by observing that all generalized H-type groups are the metric product of an ideal one (i.e.\ with no non-trivial abnormal minimizers) and a copy of the Euclidean space of the appropriate dimension.
\begin{prop}
For any generalized H-type group $G$ there exists a unique (up to isometries) ideal generalized H-type group $\hat{G}$ such that $G = \mathbb{R}^{k-2d} \times \hat{G}$, with $k-2d = \dim\ker S$.
\end{prop}
\begin{proof}
In the coordinate presentation of Section~\ref{s:normalforms}, we identify $G = \mathbb{R}^k \times \mathbb{R}^{n-k}$. According to \eqref{eq:normalformS}, we further split $G = \mathbb{R}^{k-2d} \times \mathbb{R}^{2d} \times \mathbb{R}^{n-k}$, where the first $\mathbb{R}^{k-2d}$ component represents the kernel of $S$. The sub-Riemannian structure induced by the projection on the component $\mathbb{R}^{k-2d}$ is the Euclidean one. On the other hand, the sub-Riemannian structure induced by the projection on the component $\mathbb{R}^{2d} \times \mathbb{R}^{n-k}$ is the one of a generalized H-type group of rank $2d$, with non-degenerate operator $\hat{S}$ (given by the restriction of $S$ to the orthogonal complement of $\ker S$), which we call $\hat{G}$. By Lemma~\ref{l:abnormals}, $\hat{G}$ is ideal.

By Lemma~\ref{l:geods}, all geodesics of $G$ are straight lines in the first $\mathbb{R}^{k-2d}$ component (representing the kernel of $S$), and are geodesics for $\hat{G}$ in the remaining $\mathbb{R}^{2d} \times \mathbb{R}^{n-k}$ component. This yields that $G$ is the metric product $\mathbb{R}^{k-2d} \times \hat{G}$. The uniqueness follows easily.
\end{proof}

\section{Jacobian determinant and cotangent injectivity domain}\label{s:proof}

The most important steps in the proof of Theorem~\ref{t:htype} are the computation of the Jacobian determinant of the exponential map, and the characterization of the cotangent injectivity domain of generalized H-type Carnot groups. These results have independent interests. For example, the analogous formulas for the Heisenberg group and, more generally, corank 1 Carnot groups have been employed in \cite{BKS16,BKS17} to obtain Jacobian determinant inequalities.

\addtocontents{toc}{\gobblefive} \subsection*{Notation}
For the statement of the next Lemma \ref{l:jacobian}, we recall the notation of Section \ref{s:normalforms}. For each $v \in \mathbb{R}^{n-k}$, the skew-symmetric matrix $L_v$ has real normal form given by \eqref{eq:normalformlLv}, with non-zero singular values $\alpha_i |v|$, with $i=1,\dots,d$, with $0<\alpha_1\leq \dots \leq \alpha_d$, repeated according to their multiplicity. To each $\alpha_j$ corresponds a pair of non-zero eigenvalues of the symmetric matrix $S$ defining the H-type group, see \eqref{eq:normalformS}. If $v \neq 0$, after applying an orthogonal transformation putting $L_v$ in its real normal form \eqref{eq:normalformlLv}, we denote $u=(u_0,u_1,\dots,u_d)$, where $u_0 \in \mathbb{R}^{k-2d}$ is the component in the kernel of $L_v$, while each $u_i \in \mathbb{R}^2$, for $i=1,\dots,d$, is the component of $u$ in the real eigenspace associated with the singular value $\alpha_i$. Notice that this decomposition is non-unique. The forthcoming lemma can be expressed in a more intrinsic way by introducing multiplicities of singular values, but for further manipulations we prefer to present it in its actual form.

\begin{lemma}[Jacobian determinant]\label{l:jacobian}
The Jacobian determinant of the exponential map $\exp_e(u,v) = (x,z)$ of a generalized H-type group is
\begin{multline}
J(u,v)= \frac{2^{2d}}{\alpha^2|v|^{2d+2}}\left(\sum_{i=1}^d |u_i|^2 \frac{\alpha_i|v| - \sin(\alpha_i|v|)}{2|v|^3 \alpha_i}\right)^{n-k-1} \times \\
\times \sum_{i=1}^d |u_i|^2 \prod_{j \neq i}\sin\left(\frac{\alpha_j |v|}{2}\right)^2 \sin\left(\frac{\alpha_i |v|}{2}\right)\left(\sin\left(\frac{ \alpha_j |v|}{2}\right) - \frac{\alpha_j |v|}{2} \cos\left(\frac{ \alpha_j |v|}{2}\right)\right),
\end{multline}
where $\alpha = \prod_{j=1}^d \alpha_j$ is the product of the non-zero singular values of $L_v/|v|$. If $v = 0$, the formula must be taken in the limit $v \to 0$. In particular,
\begin{equation}
J(u,0)= \left(\sum_{i=1}^d \frac{(\alpha_i |u_i|)^2}{12}\right)^{n-k} = \left(\frac{|Su|^2}{12}\right)^{n-k}.
\end{equation}
\end{lemma}
\begin{rmk}
For H-type groups, $L_v^2 = -|v|^2 \mathbbold{1}$, and the above expression reduces to
\begin{equation}
J(u,v) =  \frac{2^k|u|^{2(n-k)}}{|v|^{k+2}} \left(\frac{|v|-\sin|v|}{2|v|^3}\right)^{n-k-1} \sin\left(\frac{|v|}{2}\right)^{k-1}\left(\sin\left(\frac{|v|}{2}\right) - \frac{|v|}{2}\cos\left(\frac{|v|}{2}\right)\right).
\end{equation}
\end{rmk}
\begin{proof}
It is convenient to rewrite the exponential map $\exp_e(u,v) = (x,z)$ using polar coordinates on the $v$ and $z$ components. Namely, $v \mapsto (|v|,\hat{v} = v/|v|) \in \mathbb{R}_+ \times \mathbb{S}^{n-k-1}$, and $z \mapsto (|z|,\hat{z} = z/|z|) \in \mathbb{R}_+ \times \mathbb{S}^{n-k-1}$. One can check that $g(L_v)u \cdot u \geq 0$. Lemma~\ref{l:geods} reads
\begin{equation}\label{eq:polargeod}
x = f(L_v) u, \qquad |z| = \frac{g(L_v)u \cdot u}{2|v|}, \qquad \hat{z} = \hat{v}.
\end{equation}
Thus, passing in polar coordinates, we have
\begin{equation}\label{eq:polartocart}
J(u,v) = \frac{|z|^{n-k-1}}{|v|^{n-k-1}} \tilde{J}(u,|v|,\hat{v}),
\end{equation}
where $\tilde{J}$ is the Jacobian determinant in polar coordinates,
\begin{equation}\label{eq:protojac}
\tilde{J}(u,|v|,\hat{v}) = \det \begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial |v|} & \frac{\partial x}{\partial \hat{v}} \\
\frac{\partial |z|}{\partial u} & \frac{\partial |z|}{\partial |v|} & \frac{\partial |z|}{\partial \hat{v}} \\
\frac{\partial \hat{z}}{\partial u} & \frac{\partial \hat{z}}{\partial |v|} & \frac{\partial \hat{z}}{\partial \hat{v}} \\
\end{pmatrix} = \det
\begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial |v|} & \frac{\partial x}{\partial \hat{v}} \\
\frac{\partial |z|}{\partial u} & \frac{\partial |z|}{\partial |v|} & \frac{\partial |z|}{\partial \hat{v}} \\
\mathbbold{0} & \mathbbold{0} & \mathbbold{1} \\
\end{pmatrix}=
\det \begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial |v|} \\
\frac{\partial |z|}{\partial u} & \frac{\partial |z|}{\partial |v|}
\end{pmatrix}.
\end{equation}
Using the explicit expressions \eqref{eq:polargeod}, recalling that $L_v = |v| L_{\hat{v}}$, we obtain
\begin{align}
\frac{\partial x}{\partial u} & = f(L_v), & \frac{\partial x}{\partial |v|} & = \frac{f'(L_v) L_v}{|v|} u, \\
\frac{\partial |z|}{\partial u} & = \frac{u^*g(L_v)}{|v|}, & \frac{\partial |z|}{\partial |v|} & = \frac{[g'(L_v)L_v - g(L_v)]u \cdot u}{2|v|^2},
\end{align}
where in the third equation we used the fact that $g(L_v)^* = g(L_v^*) = g(-L_v) = g(L_v)$, the prime denotes the derivative of $f$ and $g$ defined in \eqref{eq:fg}, and where to compute the derivatives w.r.t.\ $|v|$ we exploited property \eqref{eq:derivatio}.

The matrix on the right hand side of \eqref{eq:protojac} has the following block structure
\begin{equation}
\Xi = \begin{pmatrix}
A & B \\
C & D
\end{pmatrix},
\end{equation}
with $A = f(L_v)$ a $k\times k$ matrix. If $A$ is invertible, $\det(\Xi) = \det(A)\det(D - CA^{-1} B)$. If $A$ is not invertible, we can still apply this formula replacing $A$ with a sequence of invertible matrices $A_n \to A$, and then taking the limit. We omit this standard density argument, and in the following, we will formally manipulate all the expression as if $A$ were invertible. Thus, using the fact that $f(L_v)^{-1} = (1/f)(L_v)$, we obtain
\begin{equation}
\tilde{J}(u,|v|,\hat{v}) = \frac{ \det(f(L_v))}{2|v|^2}  u^*\underbrace{\left[g'(L_v)L_v - g(L_v) - \frac{2g(L_v)f'(L_v) L_v}{f(L_v)}\right]}_{\Theta(L_v)}u.
\end{equation}
To compute the non trivial matrix $\Theta(L_v)$, we first notice that we can replace it with its symmetric part $(\Theta(L_v) + \Theta(L_v^*))/2$ without changing the result. Moreover, $\Theta(L_v)^* = \Theta(L_v^*) = \Theta(-L_v)$. Exploiting our convention, it is sufficient to perform this computation for the corresponding analytic function of one variable $\Theta(z)$. We obtain
\begin{equation}
\frac{\Theta(z) + \Theta(-z)}{2} = 2\frac{\sinh\left(\frac{z}{2}\right) - \frac{z}{2}\cosh\left(\frac{z}{2}\right)}{\sinh\left(\frac{z}{2}\right)}.
\end{equation}
Thanks to this expression, we obtain
\begin{equation}
\tilde{J}(u,|v|,\hat{v})=\frac{\det(f(L_v))}{|v|^2}u^*\left[\frac{\sinh\left(\frac{L_v}{2}\right) - \frac{L_v}{2}\cosh\left(\frac{L_v}{2}\right)}{\sinh\left(\frac{L_v}{2}\right)}\right]  u.
\end{equation}
We now perform an orthogonal change of basis in the $u$ space, in such a way that $L_v$ is in normal form \eqref{eq:normalformlLv}. Furthermore, we decompose $u=(u_0,u_1,\dots,u_d)$ according to the real eigenspaces corresponding to the kernel of $L_v$ and the pairs of eigenvalues $\pm i |v| \alpha_j$, for $j=1,\dots,d$ of $L_v$. In particular, $u_0 \in \mathbb{R}^{k-2d}$, and $u_i \in \mathbb{R}^{2}$. Then, we get
\begin{align}
\tilde{J}(u,|v|,\hat{v}) & =\frac{f(0)^{k-2d}}{|v|^2} \prod_{j=1}^d f(i\alpha_j|v|)f(-i\alpha_j|v|)
\sum_{i=1}^d |u_i|^2 \frac{\sin\left(\frac{\alpha_i|v| }{2}\right) - \frac{\alpha_i|v|}{2} \cos\left(\frac{\alpha_i|v| }{2}\right)}{\sin\left(\frac{ \alpha_i|v|}{2}\right)} \\
& = \frac{1}{|v|^2}\prod_{j=1}^d \left(\frac{2}{\alpha_j |v|} \sin\left(\frac{\alpha_j |v|}{2}\right)\right)^2
\sum_{i=1}^d |u_i|^2 \frac{\sin\left(\frac{\alpha_i |v|}{2}\right) - \frac{\alpha_i |v|}{2} \cos\left(\frac{\alpha_i |v|}{2}\right)}{\sin\left(\frac{\alpha_i |v|}{2}\right)} \\
& = \frac{2^{2d}}{\alpha^2|v|^{2d+2}}\sum_{i=1}^d |u_i|^2 \prod_{j \neq i}\sin\left(\frac{\alpha_j |v|}{2}\right)^2 \sin\left(\frac{\alpha_i |v|}{2}\right)\times \\
& \qquad\qquad\qquad\qquad\qquad\hfill\qquad\qquad \times \left(\sin\left(\frac{\alpha_j |v| }{2}\right) - \frac{\alpha_j |v|}{2} \cos\left(\frac{\alpha_j|v| }{2}\right)\right),
\end{align}
where in the second line we used the fact that $f(0) = 1$, and $\alpha = \prod_{i=1}^d \alpha_i$. From this expression, we recover the expression for the Jacobian determinant in the original Cartesian coordinates $(u,v)$ through \eqref{eq:polartocart}, and observing that
\begin{equation}
|z| = \frac{g(L_v) u \cdot u}{2|v|} = \sum_{i=1}^d |u_i|^2 \frac{\alpha_i |v| - \sin(\alpha_i |v|)}{2 \alpha_i|v|^2}. \qedhere
\end{equation}
\end{proof}

\begin{lemma}[Cotangent injectivity domain]\label{l:ingjd}
Let $\alpha_{d}$ be the largest eigenvalue of $S$, and
\begin{equation}
D := \left\lbrace \lambda = (u,v) \in T_e^*G \text{ such that } |v| < \frac{2\pi}{\alpha_d}, \text{ and }  S u \neq 0 \right\rbrace \subset T_e^*G.
\end{equation}
Then $\exp_e : D \to \exp_e(D)$ is a smooth diffeomorphism and $\mathcal{C}(e):=G \setminus \exp_e(D) $ is a closed set with zero measure. Moreover, for all $\lambda \in D$, the curve $\gamma_\lambda(t)=\exp_e(t\lambda)$, with $t \in[0,1]$, is the unique minimizing geodesic joining its endpoints.
\end{lemma}
\begin{rmk}
By Lemma~\ref{l:geods}, geodesics $\gamma_\lambda$ with $\lambda =(u,v)$ satisfying $Su=0$ or $v=0$ are Euclidean straight lines $(tu,0)$, which are optimal for all times. This observation and Lemma~\ref{l:ingjd} imply that the cut time, that is the first time of loss of optimality, is
\begin{equation}
t_{\mathrm{cut}}(u,v) = \begin{cases} \frac{2\pi}{\alpha_d|v|} & \text{if } v \neq 0 \text{ and } Su\neq 0, \\
+\infty & \text{otherwise}.
\end{cases}
\end{equation}
For standard H-type groups $S=\mathbbold{1}_{2d}$, and Remark~\ref{eq:geodesicsclassical} yields $\mathrm{Cut}_e = \{\exp(t_{\mathrm{cut}}(\lambda)\lambda)\mid \lambda \in T_e^*G\} = \{(0,z)\mid z \in \mathbb{R}^{n-k}\}$ (the center of $G$). This fact was the main result proven in \cite{Mauricio}.
\end{rmk}
\begin{proof}
Since all geodesic are normal and $(G,d)$ is complete, each point of $G$ is reached by at least one minimizing normal geodesic $\gamma_\lambda :[0,1] \to G$, with $\lambda =(u,v) \in T_e^*G$. If $|v| > 2 \pi/\alpha_d$, and $Su \neq 0$, then $\gamma_\lambda$ is not abnormal with a conjugate time at $t_* = 2 \pi |v|/\alpha_d< 1$. Since these structures are real-analytic, $\gamma_\lambda$ does not contain any abnormal segment, hence cannot be minimizing after its first conjugate time (see \cite[Chapter 8]{nostrolibro} or \cite[Appendix A]{BR-grande1}). Thus $\gamma_\lambda$ is not minimizing on $[0,1]$. On the other hand, if $Su = 0$, for any value of $v$ we obtain the same abnormal geodesic (see Lemma~\ref{l:abnormals}). It follows that $\exp_e : \bar{D} \to G$ is onto (where $\bar D$ denotes the closure of $D$). Since also $\exp_e: D \to \exp_e(D)$ is trivially onto, $\mathcal{C}(e) = G \setminus \exp_e(D) \subseteq \exp_e(\partial D)$ has zero measure.

We now prove that, if $\lambda \in D$, then $\gamma_\lambda :[0,1] \to G$ is the unique minimizing geodesic joining its endpoints. Assume by contradiction that $\lambda=(u,v)$ and $\bar\lambda=(\bar u,\bar v) \in D$, with $\lambda \neq \bar{\lambda}$, are such that $\exp_e(\lambda) = \exp_e(\bar \lambda)$. Observe that the spectrum of $L_v$ is given by
\begin{equation}\label{eq:specLv}
\sigma(L_v) = \{0,\pm i |v| \alpha_1,\dots,\pm i |v|\alpha_d\}.
\end{equation}
Using the $z$ part of Lemma~\ref{l:geods}, and \eqref{eq:specLv}, we have $g(L_v)u \cdot u / |v|^2 \geq 0$, with equality if and only if $v=0$. This readily implies that $\bar{v} = v$.
From the $x$ equation of Lemma~\ref{l:geods}, we have that $\exp_e(u,v) = \exp_e(\bar{u},v)$ is equivalent to
\begin{equation}\label{eq:kernelf}
f(L_v) (u-\bar{u}) = 0.
\end{equation}
If $v=0$, and since $f(0) = 1$, \eqref{eq:kernelf} implies $u = \bar{u}$, which is a contradiction. Hence we proceed assuming $v \neq 0$. Since $|v| < 2\pi/|\alpha_d|$, we have that $\alpha_j |v| < 2\pi$  for all $j=1,\dots,d$. It is then easy to check that $f(\pm i |v| \alpha_j) \neq 0$. This observation, \eqref{eq:kernelf} and \eqref{eq:specLv} imply that $u - \bar{u} \in \ker L_v$. Using again that $f(0) = 1$, \eqref{eq:kernelf} implies $u = \bar{u}$, which is a contradiction. Thus $\exp_e :D \to \exp_e(D)$ is a bijection.

To conclude, no $\lambda \in D$ can be a critical point for $\exp_e$, since the Jacobian determinant of Lemma~\ref{l:jacobian} is strictly positive for $|v|< 2\pi/\alpha_d$ and $Su \neq 0$.
\end{proof}

The next corollary formalizes the fact that Definition~\ref{d:MCP} for the measure contraction property makes sense in the setting of generalized H-type groups.

\begin{cor}\label{c:good}
For any $x \in G$, consider the zero measure set $\mathcal{C}(x):= \tau_x (\mathcal{C}(e))$, where $\tau_x :G \to G$ is the left-translation. There exists a measurable map $\Phi^x : G \setminus \mathcal{C}(x) \times [0,1] \to G$,
\begin{equation}
\Phi^x(y,t) := \tau_x \exp_{e}(t \exp_e^{-1}(\tau_x^{-1} y)),
\end{equation}
such that $\Phi^x(y,t)$ is the unique minimizing geodesic joining $x$ with $y$.
\end{cor}

\section{Proof of the main result}

To prove Theorem~\ref{t:htype}, we begin with a simplified version of \cite[Lemma 2.6]{Juillet}. The argument will be useful to prove further inequalities appearing in generalized H-type groups.

\begin{lemma}\label{l:ineq}
Let $g(x):= \sin(x) - x \cos(x)$. Then, for all $x \in (0,\pi)$ and $t \in [0,1]$,
\begin{equation}
g(t x) \geq t^N g(x), \qquad \forall N \geq 3.
	\end{equation}
\end{lemma}
\begin{proof}
The condition $N \geq 3$ is necessary, as $g(x) = x^3/3 + O(x^4)$. It is sufficient to prove the statement for $N=3$. The cases $t=0$ and $t=1$ are trivial, hence we assume $t \in (0,1)$. By Gronwall's Lemma the above statement is implied by the differential inequality
\begin{equation}
g'(s) \leq  3 g(s)/s, \qquad s \in (0,\pi).
\end{equation}
In fact, it is sufficient to integrate the above inequality on $[tx,x] \subset (0,\pi)$ to prove our claim. The above inequality reads
\begin{equation}
f(s):=(3-s^2) \sin (s)-3 s \cos (s) \geq 0, \qquad s \in (0, \pi).
\end{equation}
To prove it, we observe that $f(0) = 0$ and $f'(s) =  s (\sin (s)-s \cos (s)) \geq 0$ on $[0,\pi]$.
\end{proof}
The next inequality is required to deal with the extra terms appearing in the case of generalized H-type groups, due to the fact that they have corank larger than one.
\begin{lemma}\label{l:ineq2}
Let $f(x):= x - \sin(x)$. Then, for all $x \in (0,2\pi)$ and $t \in [0,1]$,
\begin{equation}
f(t x) \geq t^N f(x), \qquad \forall N \geq 3.
\end{equation}
\end{lemma}
\begin{proof}
Similarly as in the proof of Lemma~\ref{l:ineq}, and by Gronwall's Lemma applied to the interval $[tx,x] \subset (0,2\pi)$, it is sufficient to prove the differential inequality
\begin{equation}
f'(s) \leq  3 f(s)/s, \qquad s \in (0,2\pi).
\end{equation}
The above inequality reads
\begin{equation}
h(s):=2s - 3 \sin(s) + s \cos(s) \geq 0, \qquad s \in (0,2\pi).
\end{equation}
It is sufficient to observe that $h'(0)=h''(0)=0$ and $h'''(s) = s \sin(s) \geq 0$ on $[0,2\pi]$.
\end{proof}
\begin{cor}\label{c:jacobineq}
For all $(u,v) \in D$, we have the following inequality
\begin{equation}
\frac{J(t u, t v)}{J(u,v)} \geq t^{2(n-k)}, \qquad \forall t \in [0,1].
\end{equation}
\end{cor}
\begin{proof}
Apply Lemmas \ref{l:ineq} and \ref{l:ineq2} to the explicit expression of $J$ from Lemma~\ref{l:jacobian}, and then use the standard inequality $\sin(t x) \geq t \sin(x)$, valid for all $x \in [0,\pi]$ and $t \in [0,1]$.
\end{proof}

\subsection{Proof of Theorem~\ref{t:htype}}
We are now ready to prove Theorem~\ref{t:htype}.
Observe that the integer $k+3(n-k)$ coincides with the geodesic dimension of the Carnot group, by Lemma~\ref{l:fact}.

{\bf Step 1.} We first prove that the $\mathrm{MCP}(0,N)$ holds for $N\geq k+3(n-k)$. By left-translation, it is sufficient to prove the inequality \eqref{eq:MCP} for the homothety with center equal to the identity $e=(0,0)$. Let $\Omega$ be a measurable set with $0<\mu(\Omega)<+\infty$.

By Lemma~\ref{l:ingjd}, up to removing a set of zero measure, $\Omega = \exp_e(A)$ for some $A \subset D \subset T_e^*G$. On the other hand, by Corollary~\ref{c:good}, we have
\begin{equation}
\Omega_t = \exp_e(t A), \qquad \forall t \in [0,1],
\end{equation}
where $t A $ denotes the set obtained by multiplying by $t$ any point of the set $A \subset T_e^*G$ (an Euclidean homothety). Thus, for all $t \in [0,1]$ we have
\begin{align}
\mu(\Omega_t) & = \int_{\Omega_t} d \mu = \int_{t A} J(u,v) du dv \\
& = t^{n}  \int_{A} J(t u, tv) du dv \geq  t^{k+3(n-k)}  \int_{A} J(u,v) d u d v = t^{k+3(n-k)} \mu(\Omega),
\end{align}
where we used Corollary~\ref{c:jacobineq}. In particular $\mu(\Omega_t) \geq t^N \mu(\Omega)$ for all $N \geq k+3(n-k)$.

{\bf Step 2.}
Fix $\varepsilon>0$. Let $N=k+3(n-k)$. We prove that the $\mathrm{MCP}(0,N-\varepsilon)$ does not hold. Let $\lambda = (u,0) \in D$. By Lemma~\ref{l:jacobian}, and recalling that $J > 0$ on $D$, we have
\begin{equation}
J(t u,0)= t^{2(n-k)} J(u,0) < t^{2(n-k)-\varepsilon} J(u,0), \qquad \forall t \in[0,1].
\end{equation}
By continuity of $J$ and compactness of $[0,1]$, we find an open neighborhood $A \subset D$ such that $J(t\lambda) < t^{2(n-k)-\varepsilon} J(\lambda)$, for all $t \in [0,1]$ and $\lambda \in A$. Arguing as in the first step, for $\Omega= \exp_e(A)$, we obtain
\begin{equation}
\mu(\Omega_t) < t^{n+2(n-k)-\varepsilon} \mu(\Omega) = t^{N-\varepsilon}\mu(\Omega), \qquad t \in [0,1].
\end{equation}

{\bf Step 3.}
To  prove that $\mathrm{MCP}(K,N)$ does not hold for $K>0$ and any $N >1$, we observe that spaces verifying this condition are bounded (see \cite[Theorem 4.3]{Ohta-MCP}), while $G \simeq \mathbb{R}^n$ is not.
Finally, assume that $(G,d,\mu)$ satisfies $\mathrm{MCP}(K,N)$ for some $K < 0$ and $N < k+3(n-k)$. Then the scaled space $(G,\varepsilon^{-1} d, \varepsilon^{-Q} \mu)$ (where $\varepsilon>0$ and $Q = k+2(n-k)$ is the Hausdorff dimension of $(G,d)$) verifies $\mathrm{MCP}(\varepsilon^2 K,N)$ for \cite[Lemma 2.4]{Ohta-MCP}. But the two spaces $(G,d,\mu)$ and $(G,\varepsilon^{-1} d, \varepsilon^{-Q} \mu)$ are isometric through the dilation $\delta_\varepsilon(x,z):=(\varepsilon x, \varepsilon^2 z)$. In particular $(G,d,\mu)$ satisfies the $\mathrm{MCP}(\varepsilon K,N)$ for all $\varepsilon>0$, that is
\begin{equation}
\mu(\Omega_t) \geq \int_{\Omega} t \left[\frac{s_{\varepsilon K}(t d(x,z) / \sqrt{N-1})}{s_{\varepsilon K}(d(x,z) / \sqrt{N-1})}\right]^{N-1} d \mu(z), \qquad \forall t \in[0,1].
\end{equation}
Taking the limit for $\varepsilon \to 0^+$, we obtain that $(G,d,\mu)$ satisfies the $\mathrm{MCP}(0,N)$ with $N< k+3(n-k)$, but this is false as a result of the previous step.
\hfill $\qed$

redpajama_set_name: RedPajamaArXiv


To be a truly future-focused university, sustainability has to be at the heart of everything you do. At Curtin, sustainability is one of the pillars of our strategic plan and our commitment to it is evident not only through our teaching, buildings and plans for the future but through our research.
This project will expand the use of digital information modelling beyond design and construction to encompass asset management practices whilst concurrently improving sustainability, resilience and safety.
This study investigates, via life cycle assessment, a number of recycled water options for the city of Qatar and the sustainability impacts of each water supply system.
This project will map the social and affordable housing sector in Australia, to help improve the productivity, effectiveness and efficiency of the system and thus enhance access.
This study investigates the sustainability benefits associated with the reconstruction of damaged infrastructure with recycled materials instead of virgin/new construction materials.
This study models the uptake of household solar PV under various energy policy scenarios and examines the potential impacts on power utility production as households increasingly move towards increased renewable energy power sources.
This research is focused on the development of sustainability education, critical learning concepts and constructs and examines the role of sustainability threshold concepts in sustainability education.
The aim of this project is to assist Australian transport related agencies and infrastructure consultants navigate the transition to technology enabled vehicles and transport infrastructure in the coming decade.
The aim of this project is to create a carbon positive residential infill development which will have positive economic, social and environmental impacts.
The aim of this project is to use smart digital frameworks to appropriately capture the information required for key decisions made across the lifecycle of facilities in the housing, buildings, transport, utilities and airport sectors, through developing an automated integrated system for asset management.


The prompt reminded me of the Cluracan and his nemesis from The Sandman, so this snippet kind of echoes that.
I slice off a piece of my heart and lay it on the doorstep. It burns with a portion of my inner fire for a few beats, then slowly softens into a glistening lump of red muscle. Though it is no longer connected to me, it still bears traces of my life force; my Nemesis will home in on it like a fly to a recently deceased carcass.
"Did you have to do that?" says Buhawi. Buhawi's natural form is a tornado; his confrontations with his Nemesis involve a lot of shouting and fighting, and often result in the destruction of crops and manmade structures. His human manifestation is tall, bulky, and prone to anger, which is why he doesn't understand my motivations for baiting my own Nemesis. I clean my knife and put it away before attempting another explanation.
"You're a force of nature. You can afford to meet your enemies with brute strength; in fact, that's your preferred method. I, on the other hand, do not have that option." To emphasize my point, I indicate my skinny body with a sweep of my bony hand.
Buhawi snorts. "Are you saying that instead of beating your Nemesis, you're going to capture it? What's the point of that? A Nemesis is meant to be battled to the death; there's no glory in trapping it then escaping."
"I'm not going to capture it, and maybe I won't escape. I intend to have some fun with my Nemesis."
"Why do you have to make everything so complicated, Pilandok?"
My eyes twinkle as I smile. "Who says I can't have fun with my own death?"
Buhawi literally means tornado in Filipino. Pilandok is the Philippine mouse-deer, often portrayed as a trickster figure.


Lin845 (#748) 1 day 15 hours ago News land rover wing mirror All https://www.thewingmirrorcompany.co.uk Discuss Published New Discard Success!
We are focus on providing Quality wing mirrors at low prices, High Quality land wanderer driver/near side Wing Mirror Glass, land vagabond passenger/off side Wing Mirrors.
cagt (#741) 1 day 15 hours ago News uzbekistan packages uzbekistan tours All https://trendingtours.home.blog Discuss Published New Discard Success!
With this much to explore in Uzbekistan, holiday packages are an excellent approach to create the majority of your experience. Vietnam packages are a handy approach to look after whatever you need before you leave, ensuring you can merely sit back, relax.
azizulhaq (#3) 1 day 15 hours ago News geometric abstraction artsy All https://www.artsy.net Discuss Published New Discard Success!
cagt (#741) 1 day 15 hours ago News book russia tours travel packages All https://trendingtours.home.blog Discuss Published New Discard Success!
cagt (#741) 1 day 16 hours ago News moscow travel package moscow tours All https://centralasiaguidedtours.wordpress.com Discuss Published New Discard Success!
rdevanta (#15) 1 day 17 hours ago News موقع مرأة العرب الإخباري All https://thearabmirror.com Discuss Published New Discard Success!
Lin845 (#748) 1 day 17 hours ago News volkswagen wing mirror All https://www.thewingmirrorcompany.co.uk Discuss Published New Discard Success!
Bernice783 (#1446) 1 day 18 hours ago News thc All http://www.mintees.com Discuss Published New Discard Success!
One of the most common technique of drug testing in this particular country for drugs, cannabis, and also illegal materials is a urinalysis. Marijuana has been actually known to stay in one's unit for days or maybe full weeks in high enough amounts to make a good urinalysis.
seo5 (#748) 1 day 19 hours ago News agen poker terpercaya All http://www.lexispk.com Discuss Published New Discard Success!
vivekdharmani (#2791) 2 days ago News custom web page design All https://www.dharmani.com Discuss Published New Discard Success!
looking for Top Mobile App and Web Development Company? Dharmani apps is a leading web development company. Expertise in App Development globaly.
azizulhaq (#3) 2 days ago News garage repair katy tx All https://spectrumoverheaddoor.com Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News website redesign All https://wowwebsites.ca Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News web design brampton All https://wowwebsites.ca Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News website design mississauga All https://wowwebsites.ca Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News web development toronto All https://wowwebsites.ca Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News shopify partners All https://wowwebsites.ca Discuss Published New Discard Success!
wowwebsites (#748) 2 days ago News web design toronto All https://wowwebsites.ca Discuss Published New Discard Success!
ellietaylor706 (#2791) 2 days ago News gmail helpline number gmail support uk All http://www.gmailhelpline.co.uk Discuss Published New Discard Success!
Gmail is not opening Emails? If Gmail is not opening on the device then conduct an update on the application after that get your device restarted after that check the "settings" after that get the storage cleared after that get the Gmail information cleared for further information this will get the issue resolved the user will then be able to open all the Emails.
ruhikaur (#2791) 2 days ago News greater noida c@ll girls All http://www.ruhikaur.com Discuss Published New Discard Success!
Welcome to our E$corts service in Greater Noida and we wish you will have many enj0yable experiences with Independent Greater Noida E$cort. We have many clients that have frequently second-h@nd our E$corts agency for years.
elawoman12 (#251) 2 days ago News dna fragmentation All https://www.elawoman.com Discuss Published New Discard Success!