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Our inventory management systems are designed to track your documents and marketing collateral so that you will never run out of what you need. Our inventory management systems are also part of what makes our warehousing and fulfillment services so effective.
At FORMost Graphic Communications, we are adept at meeting various warehousing and fulfillment needs promptly, correctly, and efficiently. Thanks to our 30+ years of experience in the warehousing and logistics industry, we know how to store your items and fulfill your order properly. We offer same-day pick, pack, and ship services. Our dedicated inventory management services ensure that your business has the supplies that you need, and that you always know what you have..
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FORMost also offers pick and pack services. When an order is received, our warehousing professionals will gather together all of the components, place them into the proper packaging, and labeled properly for shipping. Kitting is an effective fulfillment solution for businesses looking to distribute pre-assembled items like promotional materials, samples, manuals, catalogs, media kits, product launches, PR packages, and more. By pre-building and assembling your materials, you can save a great deal of time and money in the long run. Not sure if kitting is right for you? FORMost can work with you and your business to determine what the best solution is for the products or items that you need to ship.
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This two-day 'deep-dive' Export Plan workshop is for wine businesses wanting to refine their export plan and target commercial and growth opportunities in-market.
the ins and outs of identifying, and pitching to, importers and distributors and preparing for a market visit.
Please note: for most participants, attendance at the one-day Export Ready Session is a prerequisite to attending the two-day Export Plan Workshop.


text:
Densho Digital Repository Densho Digital Repository
By Narrator
Using Materials in the DDR
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ddr-csujad-11
Family "portraits," sisters
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ddr-csujad-11-158 (CSUJAD Local ID: taz_01_57_001, CSUJAD Project ID: csudh_taz_0158)
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CSU Dominguez Hills Tazu Kawamoto Photo Album
Photographed are Tazu Kawamoto is on the left, her sister, Saye Selly Kawamoto. The caption reads "Family portraits." From page 57 of: Tazu Kawamoto photo album (csudh_taz_0001). See this object in the California State Universities Japanese American Digitization project site: taz_01_57_001
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Identity and values -- Family
Identity and values -- Nisei
Identity and values -- Women
Geographic communities -- California
CSU Dominguez Hills Department of Archives and Special Collections
[Title of item] Tazu Kawamoto Photo Album Collection, Courtesy of the Department of Archives and Special Collections. University Library. California State University, Dominguez Hills
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ddr-csujad-11-158-master-e6b2e6cf9f - Part 1
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text:
New citizens welcomed
Nineteen people swore allegiance to the Queen and were formally welcomed to the Waitaki district at yesterday's citizenship ceremony in Oamaru.
Among them was Elbert Gargar and his family from the Philippines, who came to New Zealand in 2007 to work on a dairy farm in Ngapara.
Mr Gargar said it was a very important day and a big accomplishment to have New Zealand citizenship after being in the country for seven years.
"It's a happy moment for us," he said.
Mr Gargar said to become a kiwi is a big thing as many people they work with have been so happy for them.
"In return we should be ourselves and work hard to be a good New Zealander."
After completing three seasons on the same dairy farm, Mr Gargar completed a Master of Agricultural Science in cattle fertility at Lincoln University and now works between Ashburton and Oamaru for Veehof Dairy Services as a hoof specialist.
The Gargar family also started Generation FM, a radio station based in Oamaru.
Other people to get New Zealand citizenship yesterday were – Susan Douglas (Britain), Julia Gearhart (America), Bart and Natalie Geypen (Belgium), Sofaia Leaaetoa (Tonga), Sem, Daisy, Christian and Patricia Olaguer (Philippines), Anne Harrison (Australia/Britain), Filimoemaka and Losa Kautai (Tonga), Thomas Mourtada (Czechoslovakia), Seini Ta'ufo'ouleha (Tonga).
By LINDA MCCARTHY
PHOTO: LINDA MCCARTHY
WELL PRESENTED: Dressed in the Filopino national costume is the Gargar family who were sworn in as New Zealand citizens at the Waitaki District Council yesterday. Pictured (from left) is Elbert, Forevil, Dawn, Doughbert and Douglas Gargar.
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Available on demand click here from 7pm every weekday


text:
Rifles Single-Shot
Ruger No. 1 Varminter Rifle Review
August 09, 2016 By Stan Trzoniec
Looking though my three-ring binder where I keep records of all my handloads, I took notice that my first Ruger No. 1 was purchased in February 1976. Hard to believe that was more than 40 years ago and I still have the rifle - a .22-250 Rem. in the Standard rifle configuration.
It has totaled up quite a few woodchucks for me and never failed in the field. Last year - as part of a strategy to bring out the Ruger No. 1 in a different chambering every year - Ruger reintroduced the Varminter version in .220 Swift as a distributor exclusive from Lipsey's, and this is the model I tested. This year (2016), the Varminter is offered in .243 Win.
Followers of the Ruger No. 1 will take what they can get. In years past, the No. 1 was available in six different versions and in a wide variety of chamberings. Now the selection is much more limited.
Modeled after the famous Farquharson falling-block rifle, the Ruger No. 1 was the brainchild of Bill Ruger, who fancied himself an aficionado of the single-shot rifle. Originally, the gun was to be named the Victorian, but with Bill being Bill and the gun being his No. 1 favorite, the Ruger No. 1 designation stuck.
Being a varmint model, the barrel is 26 inches in length and tapers to a pleasing 0.74 inch at the muzzle. At eight pounds, it's a little bit heavier than my Standard models but could still be considered a walking varminter.
Since this gun has a heavy barrel, there is no quarter rib and instead sports a pair of target blocks on the barrel that mate perfectly to the Ruger rings included with the gun. I like to attach a pair of Ruger rear offset rings to all my Ruger No.1 rifles to provide more wiggle room for eye relief.
Inletting on the author's sample was excellent, but as is typical on Ruger No. 1s, wood-to-metal fit in the receiver area is left proud. The polished blue finish was flawless.
Years back, this gun was deemed a "premium" model, and with that went some figured wood to go with the price tag. Today, in order to keep prices in line (over the years, the Ruger No. 1 has almost tripled in price), the wood is rather straight grained for the most part, although when I looked through the racks of a well-heeled gun store, I did turn up some above-average wood specimens.
The finish is smooth and mimics oil done in a satin patina without any dust or any other finishing blemishes. Checkering is executed in a traditional point pattern with more than ample coverage on both the pistol grip area and the fore-end.
All the inletting is first class in areas such as the tang, receiver, pistol grip cap and around the trigger guard. Around the receiver, you will find the wood is finished "proud," which simply means it is not exactly flush with the receiver flats.
Checkering is a bordered point pattern, with ample coverage on the wrist and fore-end. The wood on this particular rifle showed decent figuring.
It's not a big thing to many, and you have to remember this is a production gun, not a custom rifle. The stock is finished off with a black rubber recoil pad with a black spacer and a black pistol grip cap. Bluing and polishing on all metal parts is flawless.
While the action takes its overall look from the Farquharson rifle, it is modified and scaled-down to modern tastes. Compared to a bolt action rifle, the heart of the action measures only 2.5 inches long and only about 1.5 inches wide at its widest point. The company's design team kept the Ruger No. 1 trim by moving the mainspring forward, within the protection of the fore-end.
Upon cocking the rifle, the falling block drops down and inside the rifle. As the falling block drops down, it protrudes downward from the base of the receiver, as does the spur of the hammer, and you can feel this at the base of the operating lever just forward of the trigger guard, enabling you to tell if the gun is cocked.
The No. 1 design allows shooters to vary the strength of the ejector - kicking cases clear or easing them back so handloaders can easily capture them.
The operating lever follows the contour of the Farquharson rifle and matches the curve of the inner trigger guard, then flares downward to follow the lines of the pistol grip. There is a safety located on the tang, which makes the gun a natural for both right- and left-handed shooters.
An adjustment on the ejector strut spring allows you to choose whether you want cases ejected or, for handloaders, captured just at the safety lever.
I didn't find the rifle wanting in the accuracy department, and groups were acceptable - certainly a start in the right direction. The Swift is one of my top .22 centerfires for small game, and believe me when I tell you handloading will shrink these groups.
I have six Ruger No.1 rifles ranging from commercial to custom wildcats, and to this day, they are my No. 1 choice for varmint hunting. Maybe someday Ruger will bring this gun out in the .204 Ruger.
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The Marlin Model 1895-.444 Marlin is a handy, powerful rifle capable of taking down elk,...
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If you haven't discovered Thompson/Center's great switch-barrel gun, the Encore Pro Hunter,...
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Uberti jumps on several states' newfound acceptance of some centerfire rounds.
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2020 Carbine


This proclamation did not come as a surprise. Ever since war was declared, it was assumed that some sort of draft would be implemented.
From the flood beat: The Spokane River dropped just enough (two inches) to provide relief from the threat of flooding.
The two bridges that were imperiled the day before, and the city's water mains were spared disaster.
However, a mining man from Mullan, Idaho, went up to the summit of the Bitterroots and reported that "there were two to 10 feet of snow yet to go." He said that people should be prepared for the fact that the higher water would continue for a while.
"I shall be agreeably surprised if there is not a foot more water this year than ever before known," said the man.


03/18/2019Comments are off for this post.
Productivity is always the end game. Everyone is looking for ways to do more for less.
Today's post highlights how our franchise technology solutions counteract the top-3 productivity traps hurting big brands today.
Wasting Time on Low-Level Tasks. Not every task is worth your time and energy. For instance, if you make money selling franchises, why would you waste your time and special skills typing up financial documents? While inputting accounting data is an important job, franchisors have better things to do. That's why it's so important to delegate–or better yet, automate.
We designed our franchise technology solutions to automate as much as possible and streamline everything that needs a human touch. Our fully integrative, modular approach is inherently efficient: FranchiseSoft combines marketing, franchise development, sales, and performance functionality into a single software solution, which eliminates much of the data entry work required when working with multiple "siloed" software tools. Our solution integrates seamlessly with QuickBooks, and saves all client information for easy invoicing. Low-level support tasks are addressed using updateable FAQs and searchable training materials, so your support staff can focus on more intensive fixes that requires creative problem solving. Scheduling headaches disappear with our self-service calendar, which gives staff the ability to create and change their schedules, and immediately alerts involved parties of any conflicts or overtime hours. Even the early stages of lead generation can be automated through triggered email campaigns that welcome new prospects and move them through the sales cycle.
Burning Daylight with Excessive Back-and-Forth. Commuting back and forth from workshop to job site is a costly time sink that's also bad for the environment. But sometimes it seems unavoidable; occasionally workers need sign-off on special jobs, or you have to visit the client another day to collect payment. In these cases, rather than leaving the customer waiting for their service, you bite the bullet and do the work, knowing you'll have to make another trip in the future.
Next time you need sign-off, simply have your employee upload a picture of their work. Our franchise technology solutions let you approve of the job or point our problems remotely, while the employee is still on-site to address any issues. That way any fixes can be implemented right away.
And if you want to cut out an extra trip to collect pay, simply invoice on-site and have the client pay online.
Inefficient Training. Productivity issues can indicate larger flaws in the franchise process, but they often stem from poor training protocols. Many franchises offer little more than a week or two of training up front, which leaves plenty of room for bad habits to develop. In these situations, it's lose-lose: you either let the inefficiencies stand or backtrack on training.
That's why we designed our franchise technology solutions to make training better. In addition to your standard training process, you can create and assign custom training tasks, then monitor the trainee's progress to assess their potential and head off sticking points. FranchiseSoft's Intranet module can also be used to safely store all training materials and reference guides, which puts the answers to any problems in the palm of your employees' hands at all times.
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As with any fine jewelry item, an ultrasonic cleaner can be used. However, it is best to be gentle with your valuable treasures. Dispersion refers to the prism effect that occurs when pure white light enters a non-opaque object, breaks into spectral (rainbow) colors, and reflects back to the viewer.
Shape: Round Brilliant cut. Color: Near white. ·It will not look like a piece of rock salt, like cubic zirconia, after two weeks. Cut: Excellent.
You can be sure that we will do everything possible to make things right. in case of late information expected. all our jewelry is 100% natural conflict free. Cz Gems are Wholesale Supplier of all type of Gem Stones, precious, semiprecious stones and Rough.
SHAPE : Brilliant Round (Diamond Cut). So you can be assured of what you are buying. SIZE : (LxWxH) 6.60 x 6.60 x 3.84 MM. COLOR : White (Color Less). SI Slight Inclusion Visible At Arm's Half Length.
Color: F-G-H Full Fiery White. Very happy! Full white! Cut just like a diamond in brillance. Brillant like a diamond. Shape: Round Brilliant Cut. I'm not an expert but it looks colorless and bright.
Clarity : VVS1. Carat : 3.00 Ct. Shape : Brilliant Cushion Cut. Color : White H-I. We keep you updated regarding the status of your order.
Gem Color: White. Gem Weight: 1.00 Carat. Gem Shape: Round. Gem Quantity: 1. Item Type: Ring.


Furniture — This furniture, is one of the 13 designs published under the gallery: Elegance And ModernInterior Design By Shawn Henderson. In the next page, we have the park avenue shawnhenderson [ ... ]. While in	the previous page, we have the modern living roomflatiron shawnhenderson [ ... ].


Hi, looking to promote the launch of a web series by posting flyers around Brooklyn with a specific focus in Williamsburg and Greenpoint. Looking for someone who can pick up 100 flyers every other week over the course of 6 weeks (3 pickups total) and post them in coffeeshops, laundromats, community bulletin boards, lamp posts, etc. Need someone to also take pictures of the flyers when posted and deliver digital copies to us.


MailUp S.p.A. (the "Company" or the "Issuer" or "MailUp"), a company admitted to trading on the multilateral trading facility AIM Italia / Alternative Investment Market and operating in the marketing technology field, following the press releases respectively issued on September 19 and October 30, 2018, has announced today that the formal requirements for the final closing of Datatrics B.V.'s ("Datatrics") acquisition were completed.
In detail, with respect to the capital increase amounting to Euro 1,557,810.11 (share premium included) passed by the Board of Directors on October 30, 2018 and paid via the contribution in kind of n. 409 Datatrics shares, representing 40.94% of its share capital (the "Capital Increase in Kind"), the Issuer's Board of Directors carried out the relevant activities according to art. 2443-quater of Italian Codice Civile.
(e) the requirement for the professional capabilities and independence of the expert is met according to the applicable law.
According to art. 2343-quater, fourth paragraph, of Italian Codice Civile, since the Board of Directors resolution was registered in the Companies Register on December 13, 2018, n. 657.859 MailUp's ordinary shares, no nominal value, regular rights, deriving from the Capital Increase have been released and assigned to the relevant owners with the same ISIN as MailUp's circulating ordinary shares.
Today the declaration according to art. 2444 of Italian Codice Civile of full liberation of the Capital Increase in Kind has been registered in Milan's Company Register.


Prior to joining the No Kid Hungry Youth Ambassador program, I was not aware how widespread the issue of childhood hunger was in my community and in our country, nor was I aware of its devastating effects. I thought hunger was something that occurred in other places, to other people. But as I embarked on my Youth Ambassadorship, I quickly began to learn about and understand the negative impact hunger can have on children, families and communities. In the United States, 14.5% of families are food insecure, but in the state of Texas that number jumps to 18%, according to USDA's Household Food Security Report. Nearly 1 in 5 Texans (4.41 million) are living in poverty. As a student at Baylor University, I was compelled to do more to address this silent epidemic.


Select a rectangular area of r rows and c columns of pixels, which is completely inside the picture. As a result of the operation, each pixel inside the selected rectangle changes its colour (from black to white, or from white to black).
Initially, all pixels are white. To create a picture, the operation described above can be applied several times. Can you paint a certain picture which you have in mind?
The input contains several test cases. Each test case starts with one line containing four integers n, m, r and c. (1 ≤ r ≤ n ≤ 100, 1 ≤ c ≤ m ≤ 100), The following n lines each describe one row of pixels of the painting you want to create. The ith line consists of m characters describing the desired pixel values of the ith row in the finished painting ('0' indicates white, '1' indicates black).
The last test case is followed by a line containing four zeros.
For each test case, print the minimum number of operations needed to create the painting, or -1 if it is impossible.


text:
\section{Quadratic term in the PDW functional}\label{sec:quadratic}

The starting point is the Ginzburg-Landau (GL) functional in Eq.~(1) of the main text,
\begin{align}\label{GL}
F=\int d^3r\Bigl[&K_1\partial_k A_{\mu j}\partial_k A_{\mu j}^*
+K_2\partial_j A_{\mu j}\partial_k A_{\mu k}^*
+K_3\partial_k A_{\mu j}\partial_j A_{\mu k}^*
+\alpha \mathop{\mathrm{tr}} AA^\dag\nonumber\\
&+\beta_1|\mathop{\mathrm{tr}} AA^T|^2+\beta_2(\mathop{\mathrm{tr}} AA^\dag)^2
+\beta_3 \mathop{\mathrm{tr}} AA^T(AA^T)^*+\beta_4\mathop{\mathrm{tr}}(AA^\dag)^2
+\beta_5\mathop{\mathrm{tr}} AA^\dag(AA^\dag)^*\Bigr],
\end{align}
where $\b{r}=(x,y,z)$. We write the order parameter as
\begin{align}\label{OPexpansion}
A_{\mu j}(\b{r}_\parallel,z)=\overline{A}_{\mu j}(z)+\sum_\b{q}\phi_{\mu j}(\b{q},z)e^{i\b{q}\cdot\b{r}_\parallel},
\end{align}
where $\b{r}_\parallel=(x,y)$, $\b{q}=(q_x,q_y)$ and
\begin{align}\label{PDBOP}
\overline{A}_{\mu j}(z)=\left(\begin{array}{ccc}
\Delta_\parallel(z) & 0 & 0 \\
0 & \Delta_\parallel(z) & 0 \\
0 & 0 & \Delta_\perp(z)
\end{array}\right),
\end{align}
is the equilibrium order parameter in the planar-distorted B phase (PDB phase). We assume $\phi_{\mu j}\ll \Delta_\parallel,\Delta_\perp$ and wish to expand the GL functional to quartic order in $\phi$. The leading nontrivial term in this expansion is the term quadratic in $\phi$, which will be of the form
\begin{align}\label{FreeEnergy}
f=f[\overline{A}]+\sum_\b{q}\int_0^D dz\,\phi_{\mu j}^*(\b{q},z)\hat{C}_{\mu j,\nu k}(\b{q},z)\phi_{\nu k}(\b{q},z)+\mathcal{O}(\phi^3),
\end{align}
by translational invariance, where $f$ denotes the free energy per unit area of the film. By diagonalizing $\hat{C}$ we will be able to identity the soft mode which is relevant for the crystallization transition, as done in Ref.~\cite{mizushima2018}. We use the hatted notation to indicate that $\hat{C}$ is an operator acting on the $z$ dependence of functions to its right. Also note that since $\overline{A}$ is a stationary point of $F$, one has $\delta F/\delta{\overline{A}}=\delta F/\delta\overline{A}^*=0$ and there are no terms linear in $\phi$. To diagonalize $\hat{C}_{\mu j,\nu k}(\b{q})$, it is simplest to proceed as in $\b{k}\cdot\b{p}$ theory for the calculation of electronic bandstructures (see, e.g., Ref.~\cite{Winkler}): we will use the (``band-edge'') eigenvectors of $\hat{C}_{\mu j,\nu k}(\b{0})$ as a basis for the eigenfunctions of $\hat{C}_{\mu j,\nu k}(\b{q})$.

\subsection{Quadratic term: \texorpdfstring{$\boldsymbol{q}=0$}{G0}}\label{sec:q0}

We therefore start by assuming $\b{q}=0$, i.e., $A_{\mu j}(\b{r}_\parallel,z)=\overline{A}_{\mu j}(z)+\phi_{\mu j}(z)$. Since there is no $\b{r}_\parallel$ dependence, only gradient terms involving $\partial_z$ survive in the GL functional. First, the GL functional is invariant under the time-reversal transformation
\begin{align}
\mathcal{T}\colon A_{\mu j}\rightarrow A_{\mu j}^*.
\end{align}
Assuming $\Delta_\parallel$ and $\Delta_\perp$ real, the PDB order parameter (\ref{PDBOP}) is invariant under $\mathcal{T}$. To quadratic order, real and imaginary fluctuations cannot mix~\cite{VW3He}, thus we write $\phi_{\mu j}=\phi_{\mu j}^{(+)}+i\phi_{\mu j}^{(-)}$ with $\phi_{\mu j}^{(\pm)}$ real. Each $3\times 3$ matrix has 9 real components and can be decomposed as the sum of a trace part (proportional to the identity) and a traceless real matrix with 8 components. The latter can be decomposed into symmetric traceless (5 components) and antisymmetric (3 components) parts. A useful representation is in terms of the $SU(3)$ Gell-Mann matrices $\lambda^1,\ldots,\lambda^8$ augmented with a matrix $\lambda^0$ proportional to the identity,
\begin{align}\label{GellMann}
\lambda^0&=\sqrt{\frac{2}{3}}\left(\begin{array}{ccc}
1 & & \\
& 1 & \\
& & 1
\end{array}\right),\nonumber\\
\lambda^1&=\left(\begin{array}{ccc}
1 & & \\
& -1 & \\
& & 0
\end{array}\right),\hspace{2mm}
\lambda^2=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}
1 & & \\
& 1 & \\
& & -2
\end{array}\right),\hspace{2mm}
\lambda^3=\left(\begin{array}{ccc}
& &  \\
& & 1 \\
& 1 &
\end{array}\right),\hspace{2mm}
\lambda^4=\left(\begin{array}{ccc}
& & 1 \\
& &  \\
1 &  &
\end{array}\right),\hspace{2mm}
\lambda^5=\left(\begin{array}{ccc}
& 1 &  \\
1 & &  \\
&  &
\end{array}\right),\nonumber\\
\lambda^6&=\left(\begin{array}{ccc}
& &  \\
& & -i \\
& i &
\end{array}\right)=L_x,\hspace{2mm}
\lambda^7=\left(\begin{array}{ccc}
&  & i  \\
 & &  \\
-i &  &
\end{array}\right)=L_y,\hspace{2mm}
\lambda^8=\left(\begin{array}{ccc}
& -i &  \\
i & &  \\
&  &
\end{array}\right)=L_z,
\end{align}
and are normalized according to $\mathop{\mathrm{tr}} \lambda^I\lambda^J=2\delta^{IJ},\,I,J=0,\ldots,8$. We have also observed that $(\lambda^6,\lambda^7,\lambda^8)=(L_x,L_y,L_z)=\b{L}$ are the $J=1$ angular momentum matrices, i.e., the generators of $SO(3)$ in the vector representation. They are purely imaginary, thus we can expand
\begin{align}\label{phiExpansion}
\phi^{(\pm)}=\Delta^{(\pm)}\lambda^0+i\b{\Omega}^{(\pm)}\cdot\b{L}+\b{M}^{(\pm)}\cdot\b{\Lambda},
\end{align}
where $\Delta^{(\pm)}$ is the scalar ($J=0$) mode, $\b{\Omega}^{(\pm)}=(\Omega_1^{(\pm)},\Omega_2^{(\pm)},\Omega_3^{(\pm)})$ is the vector ($J=1$) mode, and $\b{M}^{(\pm)}=(M_1^{(\pm)},\ldots,M_5^{(\pm)})$ is the symmetric traceless tensor ($J=2$) mode, defining $\b{\Lambda}=(\lambda^1,\ldots,\lambda^5)$.

We substitute the expansion (\ref{phiExpansion}) into the GL functional and expand to quadratic order in $\phi$. We ignore the equilibrium free energy $f[\overline{A}]$, i.e., we consider the free energy relative to that of the equilibrium PDB phase. The quadratic term in the GL functional gives
\begin{align}\label{quadratic1}
f^{(2)}[\phi]=\sum_{K=\pm}\int_0^D dz\biggl\{&2K_1\left[(\partial_z\Delta^{(K)})^2+(\partial_z\b{\Omega}^{(K)})^2+(\partial_z\b{M}^{(K)})^2\right]\nonumber\\
&+K_{23}\left[\frac{2}{3}(\partial_z\Delta^{(K)}-\sqrt{2}\partial_z M_2^{(K)})^2+(\partial_z M_3^{(K)}+\partial_z\Omega_1^{(K)})^2+(\partial_z M_4^{(K)}-\partial_z\Omega_2^{(K)})^2\right]\nonumber\\
&+2\alpha\left[(\Delta^{(K)})^2+(\b{\Omega}^{(K)})^2+(\b{M}^{(K)})^2\right]\biggr\},
\end{align}
where $K_{ijk\cdots}=K_i+K_j+K_k+\cdots$. The quartic term gives
\begin{align}\label{quadratic2}
f^{(4)}[\phi]=4\sum_{K=\pm}\int_0^Ddz\biggl\{&c_\Delta^{(K)}(z)(\Delta^{(K)})^2+c_{\Omega,\parallel}^{(K)}(z)\left[(\Omega_1^{(K)})^2+(\Omega_2^{(K)})^2\right]
+c_{\Omega,\perp}^{(K)}(z)(\Omega_3^{(K)})^2+c_{M,m_J=0}^{(K)}(z)(M_2^{(K)})^2\nonumber\\
&+c_{M,m_J=\pm 1}^{(K)}(z)\left[(M_4^{(K)})^2+(M_3^{(K)})^2\right]
+c_{M,m_J=\pm 2}^{(K)}(z)\left[(M_1^{(K)})^2+(M_5^{(K)})^2\right]\nonumber\\
&+c_{M\Delta}^{(K)}(z)M_2^{(K)}\Delta^{(K)}
+c_{M\Omega}^{(K)}(z)\left(M_4^{(K)}\Omega_2^{(K)}-M_3^{(K)}\Omega_1^{(K)}\right)\biggr\},
\end{align}
where the $c$ coefficients are functions of the equilibrium order parameters $\Delta_\parallel(z)$, $\Delta_\perp(z)$ and are given by
\begin{align}
c_\Delta^{(+)}(z)&=\frac{1}{3}\left[2(7\beta_{12}+3\beta_{345})\Delta_\parallel^2+8\beta_{12}\Delta_\parallel\Delta_\perp+(5\beta_{12}+3\beta_{345})\Delta_\perp^2\right],\\
c_\Delta^{(-)}(z)&=\frac{1}{3}\left[2(\beta_1+3\beta_2+\beta_{345})\Delta_\parallel^2+8\beta_1\Delta_\parallel\Delta_\perp-(\beta_1-3\beta_2-\beta_{345})\Delta_\perp^2\right],
\end{align}
for terms involving the $J=0$ modes $\Delta^{(\pm)}$ alone,
\begin{align}
c_{\Omega,\parallel}^{(+)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2-\beta_{345}\Delta_\parallel\Delta_\perp+\beta_{12345}\Delta_\perp^2,\\
c_{\Omega,\parallel}^{(-)}(z)&=-(2\beta_1-2\beta_2-\beta_4)\Delta_\parallel^2-(\beta_{35}-\beta_4)\Delta_\parallel\Delta_\perp-(\beta_1-\beta_2-\beta_4)\Delta_\perp^2,\\
c_{\Omega,\perp}^{(+)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2+\beta_{12}\Delta_\perp^2,\\
c_{\Omega,\perp}^{(-)}(z)&=-(2\beta_1-2\beta_2+\beta_{35}-3\beta_4)\Delta_\parallel^2-(\beta_1-\beta_2)\Delta_\perp^2,
\end{align}
for terms involving the $J=1$ modes $\b{\Omega}^{(\pm)}$ alone,
\begin{align}
c_{M,m_J=0}^{(+)}(z)&=\frac{1}{3}\left[(10\beta_{12}+3\beta_{345})\Delta_\parallel^2-8\beta_{12}\Delta_\parallel\Delta_\perp+(7\beta_{12}+6\beta_{345})\Delta_\perp^2\right],\\
c_{M,m_J=0}^{(-)}(z)&=-\frac{1}{3}\left[(2\beta_1-6\beta_2-\beta_{345})\Delta_\parallel^2+8\beta_1\Delta_\parallel\Delta_\perp-(\beta_1+3\beta_2+2\beta_{345})\Delta_\perp^2\right],\\
c_{M,m_J=\pm 1}^{(+)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2+\beta_{345}\Delta_\parallel\Delta_\perp+\beta_{12345}\Delta_\perp^2,\\
c_{M,m_J=\pm 1}^{(-)}(z)&=-(2\beta_1-2\beta_2-\beta_4)\Delta_\parallel^2-(\beta_4-\beta_{35})\Delta_\parallel\Delta_\perp-(\beta_1-\beta_{24})\Delta_\perp^2,\\
c_{M,m_J=\pm 2}^{(+)}(z)
&=(2\beta_{12}+3\beta_{345})\Delta_\parallel^2+\beta_{12}\Delta_\perp^2,\\
c_{M,m_J=\pm 2}^{(-)}(z)&=-(2\beta_1-2\beta_2-\beta_{345})\Delta_\parallel^2-(\beta_1-\beta_2)\Delta_\perp^2,
\end{align}
for terms involving the $J=2$ modes $\b{M}^{(\pm)}$ alone, and
\begin{align}
c_{M\Delta}^{(+)}(z)&=\frac{2\sqrt{2}}{3}(\Delta_\parallel-\Delta_\perp)\left[(4\beta_{12}+3\beta_{345})\Delta_\parallel+(2\beta_{12}+3\beta_{345})\Delta_\perp\right],\\
c_{M\Delta}^{(-)}(z)&=\frac{2\sqrt{2}}{3}(\Delta_\parallel-\Delta_\perp)\left[(4\beta_1+\beta_{345})\Delta_\parallel+(2\beta_1+\beta_{345})\Delta_\perp\right],\\
c_{M\Omega}^{(+)}(z)&=0,\\
c_{M\Omega}^{(-)}(z)&=2(\beta_3-\beta_5)(\Delta_\parallel^2-\Delta_\perp^2),
\end{align}
for couplings between modes. We use the notation $\beta_{ijk\cdots}=\beta_i+\beta_j+\beta_k+\cdots$.

The reduced functional (\ref{quadratic1}-\ref{quadratic2}) should preserve the $SO(2)_{L_z+S_z}$ symmetry of the PDB phase. To determine how $\phi$ transforms under this symmetry, first recall the action of the residual $SO(3)_{L+S}$ symmetry in the bulk B phase~\cite{VW3He}:
\begin{align}\label{SOrotation}
A_{\mu j}\rightarrow R_{\mu\nu}^{(S)}R_{jk}^{(L)}A_{\nu k}=\left(R^{(S)}AR^{(L)T}\right)_{\mu j},
\end{align}
where $A=e^{i\phi}R$, $R^{(L)}=O$, and $R^{(S)}=ROR^{-1}$, with $R,O\in SO(3)$. By relative spin-orbit symmetry, one can choose $R=\mathbb{I}$, in which case the residual symmetry acts by conjugation, $A\rightarrow OAO^{-1}$. For a $J_z$ rotation about the $z$ axis through angle $\theta$, one has $O=e^{-i\theta L_z}$ with $L_z=\lambda^8$ defined earlier. This is clearly a symmetry of the PDB phase since $e^{-i\theta L_z}\overline{A}e^{i\theta L_z}=\overline{A}$. Fluctuations will transform as $\phi\rightarrow e^{-i\theta L_z}\phi e^{i\theta L_z}$; it is sufficient to consider infinitesimal rotations, $\phi\rightarrow \phi+\delta\phi$ where $\delta\phi=-i\theta[L_z,\phi]$. Since $L_z$ is purely imaginary, this transformation does not mix real and imaginary modes (it commutes with time reversal, as expected). One can thus write the first variation as $\delta\phi=\delta\phi^{(+)}+i\delta\phi^{(-)}$ where $\delta\phi^{(\pm)}$ are real, and can thus be expanded as in (\ref{phiExpansion}):
\begin{align}
\delta\phi^{(\pm)}=\delta\Delta^{(\pm)}\lambda^0+i\delta\b{\Omega}^{(\pm)}\cdot\b{L}+\delta\b{M}^{(\pm)}\cdot\b{\Lambda}.
\end{align}
On the other hand, we have
\begin{align}
\delta\phi^{(\pm)}=-i\theta[L_z,\phi^{(\pm)}]
=-i\theta\epsilon_{ij}\Omega_j^{(\pm)}L_i-i\theta M_a^{(\pm)}[\lambda^8,\lambda^a],
\end{align}
where $i,j=1,2$ and $a=1,\ldots,5$, and we have used the commutation relations for the angular momentum operators. We thus find
\begin{align}\label{JzTrafo1}
\delta\Delta^{(\pm)}=0;\hspace{5mm}
\delta\Omega_3^{(\pm)}=0;\hspace{5mm}
\delta\Omega_i^{(\pm)}=-\theta\epsilon_{ij}\Omega_j^{(\pm)},\,i,j=1,2.
\end{align}
This is as expected: $\Delta^{(\pm)}$ is a scalar mode that does not transform under rotations; $\Omega_3^{(\pm)}$ is the $z$ component of a vector mode, that is left invariant under $J_z$ rotations; and $\Omega_1^{(\pm)}$ and $\Omega_2^{(\pm)}$ are the $x$ and $y$ components of this vector mode, which transform into each other under $J_z$ rotations.

Turning to the $J=2$ modes, one can check that $i[\lambda^8,\lambda^a]$ is real symmetric traceless, thus it can be expanded in the basis of $\lambda^a$ matrices:
\begin{align}
-i[\lambda^8,\lambda^a]=\sum_{b=1}^5\mathcal{J}_{ab}\lambda^b,\,a=1,\ldots,5,
\end{align}
with $\mathcal{J}$ a real antisymmetric $5\times 5$ matrix. Using the orthogonality property of Gell-Mann matrices, we have \begin{align}
\mathcal{J}_{ab}=\frac{1}{2i}\mathop{\mathrm{tr}}([\lambda^8,\lambda^a]\lambda^b)=\left(\begin{array}{ccccc}
0 & 0 & 0 & 0 & 2 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
-2 & 0 & 0 & 0 & 0
\end{array}\right)_{ab}.
\end{align}
Thus we find $\delta M_a^{(\pm)}=-\theta\mathcal{J}_{ab}M_b^{(\pm)}$. More specifically, we have
\begin{align}\label{JzTrafo2}
\delta M_2^{(\pm)}&=0;\nonumber\\
\delta M_4^{(\pm)}&=-\theta M_3^{(\pm)},\hspace{5mm}
\delta M_3^{(\pm)}=\theta M_4^{(\pm)};\nonumber\\
\delta M_1^{(\pm)}&=-2\theta M_5^{(\pm)},\hspace{5mm}
\delta M_5^{(\pm)}=2\theta M_1^{(\pm)}.
\end{align}
Once again, there is a clear interpretation: $M_2^{(\pm)}$ is an $m_J=0$ mode (like a $d_{3z^2-r^2}$ orbital), invariant under $J_z$ rotations; $M_4^{(\pm)}$ and $M_3^{(\pm)}$ can be combined to form $m_J=\pm 1$ modes (like $d_{xz}\pm i d_{yz}$ orbitals); and $M_1^{(\pm)}$ and $M_5^{(\pm)}$ can be combined to form $m_J=\pm 2$ modes (like $d_{x^2-y^2}\pm i d_{xy}$ orbitals). Using the transformation properties above, one can check explicitly that the functional (\ref{quadratic1}-\ref{quadratic2}) is invariant under infinitesimal $SO(2)_{J_z}$ rotations.

In order to find the normal modes, we will first bring the gradient terms in Eq.~(\ref{quadratic1}) to diagonal form by performing an orthogonal transformation. In terms of the rotated fields
\begin{align}\label{RotatedFields1}
\tilde{\Delta}^{(K)}=\frac{1}{\sqrt{3}}(\Delta^{(K)}-\sqrt{2}M_2^{(K)}),\hspace{5mm}
\tilde{\b\Omega}^{(K)}=\left(\begin{array}{c}
\tilde{\Omega}_1^{(K)} \\
\tilde{\Omega}_2^{(K)} \\
\tilde{\Omega}_3^{(K)}
\end{array}\right)=
\left(\begin{array}{c}
\frac{1}{\sqrt{2}}(\Omega_1^{(K)}-M_3^{(K)}) \\
\frac{1}{\sqrt{2}}(\Omega_2^{(K)}+M_4^{(K)}) \\
\Omega_3^{(K)}
\end{array}\right),
\end{align}
and
\begin{align}\label{RotatedFields2}
\tilde{\b{M}}^{(K)}=\left(\begin{array}{c}
\tilde{M}_1^{(K)} \\
\tilde{M}_2^{(K)} \\
\tilde{M}_3^{(K)} \\
\tilde{M}_4^{(K)} \\
\tilde{M}_5^{(K)}
\end{array}\right)
=\left(\begin{array}{c}
M_1^{(K)} \\
\frac{1}{\sqrt{3}}(M_2^{(K)}+\sqrt{2}\Delta^{(K)}) \\
\frac{1}{\sqrt{2}}(M_3^{(K)}+\Omega_1^{(K)}) \\
\frac{1}{\sqrt{2}}(M_4^{(K)}-\Omega_2^{(K)}) \\
M_5^{(K)}
\end{array}\right),
\end{align}
Equation~(\ref{quadratic1}) becomes
\begin{align}
f^{(2)}[\phi]=\sum_{K=\pm}\int_0^Ddz\Bigl\{&2K_{123}\left[(\partial_z\tilde{\Delta}^{(K)})^2
+(\partial_z\tilde{M}_3^{(K)})^2+(\partial_z\tilde{M}_4^{(K)})^2\right]\nonumber\\
&+2K_1\left[(\partial_z\tilde{\b{\Omega}}^{(K)})^2+(\partial_z\tilde{M}_2^{(K)})^2+(\partial_z\tilde M_1^{(K)})^2+(\partial_z\tilde M_5^{(K)})^2\right]\nonumber\\
&+2\alpha\left[(\tilde{\Delta}^{(K)})^2+(\tilde{\b{\Omega}}^{(K)})^2+(\tilde{\b{M}}^{(K)})^2\right]\Bigr\},
\end{align}
and Eq.~(\ref{quadratic2}) assumes the same form but with different (tilded) coefficients:
\begin{align}
f^{(4)}[\phi]=4\sum_{K=\pm}\int_0^Ddz\biggl\{&\tilde c_\Delta^{(K)}(z)(\tilde\Delta^{(K)})^2+\tilde c_{\Omega,\parallel}^{(K)}(z)\left[(\tilde\Omega_1^{(K)})^2+(\tilde\Omega_2^{(K)})^2\right]
+\tilde c_{\Omega,\perp}^{(K)}(z)(\tilde\Omega_3^{(K)})^2+\tilde c_{M,m_J=0}^{(K)}(z)(\tilde M_2^{(K)})^2\nonumber\\
&+\tilde c_{M,m_J=\pm 1}^{(K)}(z)\left[(\tilde M_4^{(K)})^2+(\tilde M_3^{(K)})^2\right]
+\tilde c_{M,m_J=\pm 2}^{(K)}(z)\left[(\tilde M_1^{(K)})^2+(\tilde M_5^{(K)})^2\right]\nonumber\\
&+\tilde c_{M\Delta}^{(K)}(z)\tilde M_2^{(K)}\tilde\Delta^{(K)}
+\tilde c_{M\Omega}^{(K)}(z)\left(\tilde M_4^{(K)}\tilde\Omega_2^{(K)}-\tilde M_3^{(K)}\tilde\Omega_1^{(K)}\right)\biggr\},
\end{align}
with
\begin{align}
\tilde c_\Delta^{(+)}(z)&=2\beta_{12}\Delta_\parallel^2+3(\beta_{12}+\beta_{345})\Delta_\perp^2,\\
\tilde c_\Delta^{(-)}(z)&=-2(\beta_1-\beta_2)\Delta_\parallel^2+\beta_{12345}\Delta_\perp^2,
\end{align}

\begin{align}
\tilde c_{\Omega,\parallel}^{(+)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2+(\beta_{12}+\beta_{345})\Delta_\perp^2,\\
\tilde c_{\Omega,\parallel}^{(-)}(z)&=-(2\beta_1-2\beta_2-\beta_3-\beta_4+\beta_5)\Delta_\parallel^2-(\beta_1-\beta_2+\beta_3-\beta_4-\beta_5)\Delta_\perp^2,\\
\tilde c_{\Omega,\perp}^{(+)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2+\beta_{12}\Delta_\perp^2,\\
\tilde c_{\Omega,\perp}^{(-)}(z)&=-(2\beta_1-2\beta_2+\beta_3-3\beta_4+\beta_5)\Delta_\parallel^2-(\beta_1-\beta_2)\Delta_\perp^2,
\end{align}

\begin{align}
\tilde c_{M,m_J=0}^{(+)}(z)&=3(2\beta_{12}+\beta_{345})\Delta_\parallel^2+\beta_{12}\Delta_\perp^2,\\
\tilde c_{M,m_J=0}^{(-)}(z)&=(2\beta_{12}+\beta_{345})\Delta_\parallel^2-(\beta_1-\beta_2)\Delta_\perp^2,\\
\tilde c_{M,m_J=\pm 1}^{(+)}(z)&=\tilde c_{\Omega,\parallel}^{(+)}(z)=(2\beta_{12}+\beta_{345})\Delta_\parallel^2+(\beta_{12}+\beta_{345})\Delta_\perp^2,\\
\tilde c_{M,m_J=\pm 1}^{(-)}(z)&=-(2\beta_1-2\beta_2+\beta_3-\beta_4-\beta_5)\Delta_\parallel^2-(\beta_1-\beta_2-\beta_3-\beta_4+\beta_5)\Delta_\perp^2,\\
\tilde c_{M,m_J=\pm 2}^{(+)}(z)&=(2\beta_{12}+3\beta_{345})\Delta_\parallel^2+\beta_{12}\Delta_\perp^2,\\
\tilde c_{M,m_J=\pm 2}^{(-)}(z)&=-(2\beta_1-2\beta_2-\beta_{345})\Delta_\parallel^2-(\beta_1-\beta_2)\Delta_\perp^2,
\end{align}

\begin{align}
\tilde c_{M\Delta}^{(+)}(z)&=4\sqrt{2}\beta_{12}\Delta_\parallel\Delta_\perp,\\
\tilde c_{M\Delta}^{(-)}(z)&=4\sqrt{2}\beta_1\Delta_\parallel\Delta_\perp,\\
\tilde c_{M\Omega}^{(+)}(z)&=2\beta_{345}\Delta_\parallel\Delta_\perp,\\
\tilde c_{M\Omega}^{(-)}(z)&=2(\beta_3-\beta_4+\beta_5)\Delta_\parallel\Delta_\perp.
\end{align}

Using the definitions (\ref{RotatedFields1}-\ref{RotatedFields2}) and the transformation properties (\ref{JzTrafo1}) and (\ref{JzTrafo2}), we can show that the tilded fields transform under $SO(2)_{J_z}$ rotations in the same way as the old ones, i.e.,
\begin{align}
\delta\tilde{\Delta}^{(K)}&=\delta\tilde{\Omega}_3^{(K)}=\delta\tilde{M}_2^{(K)}=0,\\
\delta\tilde\Omega_1^{(K)}&=-\theta\tilde\Omega_2^{(K)},\hspace{5mm}\delta\tilde\Omega_2^{(K)}=\theta\tilde\Omega_1^{(K)},\\
\delta\tilde M_4^{(K)}&=-\theta\tilde M_3^{(K)},\hspace{5mm}\delta\tilde M_3^{(K)}=\theta\tilde M_4^{(K)},\\
\delta\tilde M_1^{(K)}&=-2\theta\tilde M_5^{(K)},\hspace{5mm}\delta\tilde M_5^{(K)}=2\theta\tilde M_1^{(K)}.
\end{align}

The functional can be written in block-diagonal form. Schematically, we have:
\begin{align}
\{\tilde\Delta,\tilde M_2\}\oplus\{\tilde\Omega_1,\tilde M_3\}\oplus\{\tilde\Omega_2,\tilde M_4\}\oplus\tilde\Omega_3\oplus\tilde M_1\oplus\tilde M_5.
\end{align}
We begin with the $J=0,2$, $m_J=0$ block $\{\tilde\Delta,\tilde M_2\}$. We consider specular boundary conditions at $z=0$ and $z=D$, and the tilded quantities obey the same (Neumann or Dirichlet) boundary conditions as $\overline{A}$. Integration by parts thus gives
\begin{align}
\int_0^Ddz\,(\partial_z \tilde{g})^2=-\int_0^D dz\,\tilde{g}\partial_z^2 \tilde{g},
\end{align}
for a function $\tilde{g}$ satisfying such boundary conditions. Using the notation $f_{J;m_J}$ for each block's contribution to the GL functional and $\hat\Pi^{(K)}_{J;m_J}$ for the corresponding quadratic kernel, we have
\begin{align}
f_{0,2;0}=\sum_{K=\pm}\int_0^D dz\,\left(\begin{array}{c}
\tilde{\Delta}^{(K)} \\
\tilde{M}_2^{(K)}
\end{array}\right)^T
\hat\Pi_{0,2;0}^{(K)}
\left(\begin{array}{c}
\tilde{\Delta}^{(K)} \\
\tilde{M}_2^{(K)}
\end{array}\right),
\end{align}
where
\begin{align}\label{PiJ020}
\left(\hat\Pi_{0,2;0}^{(K)}\right)_{11}&=-2K_{123}\partial_z^2+2\alpha+4\tilde{c}_{\Delta}^{(K)}(z),\nonumber\\
\left(\hat\Pi_{0,2;0}^{(K)}\right)_{12}&=\left(\hat\Pi_{0,2;0}^{(K)}\right)_{21}=2\tilde{c}_{M\Delta}^{(K)}(z),\nonumber\\
\left(\hat\Pi_{0,2;0}^{(K)}\right)_{22}&=-2K_1\partial_z^2+2\alpha+4\tilde{c}_{M,m_J=0}^{(K)}(z).
\end{align}
Using these expressions we can identify the $U(1)$ (phase) Goldstone mode. This corresponds to pure phase fluctuations of the order parameter:
\begin{align}
A(\b{r})=\overline{A}(z)e^{i\varphi(\b{r})}\approx\overline{A}(z)(1+i\varphi),
\end{align}
thus $\phi_{U(1)}=i\overline{A}\varphi$, implying $\phi^{(+)}_{U(1)}=0$ and $\phi^{(-)}_{U(1)}=\overline{A}\varphi$. Goldstone's theorem implies that uniform fluctuations, i.e., $\varphi=\text{const.}$, cost zero energy. To verify this, we need to show that $\hat{\Pi}_{0,2;0}^{(-)}$ acting on a fluctuation of this type gives zero. Decomposing $\phi^{(-)}_{U(1)}$ as in Eq.~(\ref{phiExpansion}), and using Eqs.~(\ref{RotatedFields1}-\ref{RotatedFields2}) to express the result in terms of the tilded fields, we find
\begin{align}
\tilde{\Delta}_{U(1)}^{(-)}=\frac{\varphi}{\sqrt{2}}\Delta_\perp,\hspace{5mm}
\tilde{M}_{2,U(1)}^{(-)}=\varphi\Delta_\parallel.
\end{align}
(Since $\phi^{(-)}_{U(1)}$ is diagonal, only $\lambda^0$, $\lambda^1$, and $\lambda^2$ may contribute; since $\overline{A}_{xx}=\overline{A}_{yy}$, there is in fact no $\lambda^1$ component.) Using Eq.~(\ref{PiJ020}), we find
\begin{align}\label{GoldstoneCheckJ020}
\hat\Pi_{0,2;0}^{(-)}
\left(\begin{array}{c}
\tilde{\Delta}^{(-)}_{U(1)} \\
\tilde{M}_{2,U(1)}^{(-)}
\end{array}\right)=\sqrt{2}\varphi\left(
\begin{array}{c}
\left[-K_{123}\partial_z^2+\alpha+2\beta_{12345}\Delta_\perp^2+4\beta_{12}\Delta_\parallel^2\right]\Delta_\perp \\
\sqrt{2}\left[-K_1\partial_z^2+\alpha+2(2\beta_{12}+\beta_{345})\Delta_\parallel^2+2\beta_{12}\Delta_\perp^2\right]\Delta_\parallel
\end{array}\right).
\end{align}
On the other hand, one verifies that the GL equations $\delta f/\delta\overline{A}=0$ for $\Delta_\parallel(z)$ and $\Delta_\perp(z)$ are precisely
\begin{align}
\left[-K_1\partial_z^2+\alpha+2(2\beta_{12}+\beta_{345})\Delta_\parallel^2+2\beta_{12}\Delta_\perp^2\right]\Delta_\parallel&=0,\\
\left[-K_{123}\partial_z^2+\alpha+2\beta_{12345}\Delta_\perp^2+4\beta_{12}\Delta_\parallel^2\right]\Delta_\perp&=0,
\end{align}
thus the right-hand side of Eq.~(\ref{GoldstoneCheckJ020}) identically vanishes, provided that the order parameter profiles $\Delta_\parallel(z)$ and $\Delta_\perp(z)$ satisfy the exact GL equations.

Besides the $U(1)$ Goldstone mode, from the symmetry-breaking pattern $SO(3)_S\times SO(2)_{L_z}\rightarrow SO(2)_{L_z+S_z}$ for the PDB phase in a slab geometry we expect 3 more Goldstone modes, 2 corresponding to ordinary spin waves (broken $S_x$ and $S_y)$ rotations and 1 corresponding to a spin-orbit mode (broken relative $L_z-S_z$ rotations). Beginning with the spin waves, they correspond to pure spin rotations about the $\hat{\b{x}}$ and $\hat{\b{y}}$ axes:
\begin{align}
A_{\mu j}(\b{r})=R^{(S)}_{\mu\nu}(\hat{\b{x}},\varphi(\b{r}))\overline{A}_{\nu j}(z)\hspace{5mm}
\text{or}\hspace{5mm}
A_{\mu j}(\b{r})=R^{(S)}_{\mu\nu}(\hat{\b{y}},\varphi(\b{r}))\overline{A}_{\nu j}(z),
\end{align}
with $R^{(S)}(\hat{\b{x}},\varphi)=e^{-i\varphi L_x}$ and $R^{(S)}(\hat{\b{y}},\varphi)=e^{-i\varphi L_y}$. Expanding for small $\varphi$, we find that these modes are real with $\phi_{S_x}^{(+)}=-i\varphi L_x\overline{A}$ and $\phi_{S_y}^{(+)}=-i\varphi L_y\overline{A}$. Decomposing in the Gell-Mann basis and rotating the fields, we find that only the $m_J=\pm 1$ modes $\tilde{\Omega}_{1,2}$ and $\tilde{M}_{3,4}$ are involved:
\begin{align}
\tilde{\Omega}_{1,S_x}^{(+)}&=-\frac{\varphi}{\sqrt{2}}\Delta_\parallel,\hspace{5mm}
\tilde{M}_{3,S_x}^{(+)}=-\frac{\varphi}{\sqrt{2}}\Delta_\perp;\\
\tilde{\Omega}_{2,S_y}^{(+)}&=-\frac{\varphi}{\sqrt{2}}\Delta_\parallel,\hspace{5mm}
\tilde{M}_{4,S_y}^{(+)}=\frac{\varphi}{\sqrt{2}}\Delta_\perp,
\end{align}
with all other components zero. The $m_J=\pm 1$ fluctuation free energy is
\begin{align}
f_{1,2;\pm 1}=\sum_{K=\pm}\int_0^D dz\left[\left(\begin{array}{c}
\tilde{\Omega}_1^{(K)} \\
\tilde{M}_3^{(K)}
\end{array}\right)^T
\hat\Pi_{1,2;x}^{(K)}
\left(\begin{array}{c}
\tilde{\Omega}_1^{(K)} \\
\tilde{M}_3^{(K)}
\end{array}\right)
+
\left(\begin{array}{c}
\tilde{\Omega}_2^{(K)} \\
\tilde{M}_4^{(K)}
\end{array}\right)^T
\hat\Pi_{1,2;y}^{(K)}
\left(\begin{array}{c}
\tilde{\Omega}_2^{(K)} \\
\tilde{M}_4^{(K)}
\end{array}\right)\right],
\end{align}
where
\begin{align}
\left(\hat\Pi_{1,2;x}^{(K)}\right)_{11}&=-2K_1\partial_z^2+2\alpha+4\tilde c_{\Omega,\parallel}^{(K)}(z),\nonumber\\
\left(\hat\Pi_{1,2;x}^{(K)}\right)_{12}&=\left(\hat\Pi_{1,2;x}^{(K)}\right)_{21}=-2\tilde c_{M\Omega}^{(K)}(z),\nonumber\\
\left(\hat\Pi_{1,2;x}^{(K)}\right)_{22}&=-2K_{123}\partial_z^2+2\alpha+4\tilde c_{M,m_J=\pm 1}^{(K)}(z),
\end{align}
and
\begin{align}
\left(\hat\Pi_{1,2;y}^{(K)}\right)_{11}&=\left(\hat\Pi_{1,2;x}^{(K)}\right)_{11},\nonumber\\
\left(\hat\Pi_{1,2;y}^{(K)}\right)_{12}&=\left(\hat\Pi_{1,2;y}^{(K)}\right)_{21}=-\left(\hat\Pi_{1,2;x}^{(K)}\right)_{12},\nonumber\\
\left(\hat\Pi_{1,2;y}^{(K)}\right)_{22}&=\left(\hat\Pi_{1,2;x}^{(K)}\right)_{22}.
\end{align}
One straightforwardly verifies that a uniform $S_x$ or $S_y$ spin wave (i.e., $\varphi=\text{const.}$) costs zero energy, i.e.,
\begin{align}
\hat\Pi_{1,2;x}^{(+)}\left(\begin{array}{c}
\tilde{\Omega}_{1,S_x}^{(+)} \\
\tilde{M}_{3,S_x}^{(+)}
\end{array}\right)=0,\hspace{5mm}
\hat\Pi_{1,2;y}^{(+)}\left(\begin{array}{c}
\tilde{\Omega}_{2,S_y}^{(+)} \\
\tilde{M}_{4,S_y}^{(+)}
\end{array}\right)=0,
\end{align}
if $\Delta_\parallel(z)$ and $\Delta_\perp(z)$ satisfy the GL equations.

Finally, we turn to the Goldstone mode associated with the breaking $SO(2)_{S_z}\times SO(2)_{L_z}\rightarrow SO(2)_{L_z+S_z}$ of relative spin-orbit symmetry. This mode corresponds to an $L_z$ rotation by ${\textstyle{\frac{1}{2}}}\varphi(\b{r})$ and a simultaneous $S_z$ rotation by $-{\textstyle{\frac{1}{2}}}\varphi(\b{r})$:
\begin{align}
A_{\mu j}(\b{r})=R_{\mu\nu}^{(S)}(\hat{\b{z}},-{\textstyle{\frac{1}{2}}}\varphi(\b{r}))R_{jk}^{(L)}(\hat{\b{z}},{\textstyle{\frac{1}{2}}}\varphi(\b{r}))\overline{A}_{\nu k}(z).
\end{align}
Expanding for small $\varphi$, we find that this mode is real with $\phi^{(+)}_\text{SO}=\frac{i\varphi}{2}\{L_z,\overline{A}\}=i\varphi\Delta_\parallel L_z$. Thus only the $J=1$, $m_J=0$ component $\Omega_3$ is nonzero, and
\begin{align}
\tilde{\Omega}_{3,\text{SO}}^{(+)}=\varphi\Delta_\parallel.
\end{align}
The $J=1$, $m_J=0$ fluctuation free energy is
\begin{align}
f_{1;0}=\sum_{K=\pm}\int_0^D dz\,\tilde{\Omega}_3^{(K)}\hat{\Pi}_{1;0}^{(K)}\tilde{\Omega}_3^{(K)},
\end{align}
where
\begin{align}
\hat{\Pi}_{1;0}^{(K)}=-2K_1\partial_z^2+2\alpha+4\tilde c_{\Omega,\perp}^{(K)}(z).
\end{align}
One again straightforwardly checks that $\hat{\Pi}_{1;0}^{(+)}\tilde{\Omega}_{3,\text{SO}}^{(+)}=0$ for a uniform fluctuation $\varphi=\text{const.}$, provided that $\Delta_\parallel(z)$ and $\Delta_\perp(z)$ satisfy the GL equations, which identifies $\tilde{\Omega}_{3,\text{SO}}^{(+)}$ as the spin-orbit Goldstone mode.

The remaining contribution to the fluctuation free energy is that of the $J=2$, $m_J=\pm 2$ modes $\tilde{M}_1$ and $\tilde{M}_5$:
\begin{align}
f_{2;\pm 2}=\sum_{K=\pm}\sum_{a=1,5}\int_0^D dz\,\tilde{M}_a^{(K)}\hat\Pi_{2;\pm 2}^{(K)}\tilde{M}_a^{(K)},
\end{align}
where
\begin{align}
\hat\Pi_{2;\pm 2}^{(K)}=-2K_1\partial_z^2+2\alpha+4\tilde c_{M,m_J=\pm 2}^{(K)}(z).
\end{align}
Since we have exhausted the number (4) of Goldstone modes, such modes are necessarily massive.

\subsection{Quadratic term: \texorpdfstring{$\boldsymbol{q}\neq 0$}{Gneq0}}\label{sec:qneq0}

Mizushima and Sauls~\cite{mizushima2018} find that the mode that goes soft at the superfluid crystallization transition is an $m=0^+$ mode, which in our notation belongs to the $J=0,2$, $m_J=0$ multiplet $\{\tilde{\Delta}^{(+)},\tilde{M}_2^{(+)}\}$. We thus focus on the $K=+$ modes in what follows. Having developed the formalism for the ``band-edge'' modes with $\b{q}=0$, we return to the full free energy (\ref{FreeEnergy}), which contains a sum over all in-plane wavevectors $\b{q}$. The real and imaginary modes do not mix at quadratic order, even with $\b{q}\neq 0$. Ignoring the equilibrium piece $f[\overline{A}]$ from now on, Eq.~(\ref{FreeEnergy}) becomes:
\begin{align}\label{FreeEnergyQuad}
f=\sum_\b{q}\int_0^D dz\,\Phi_{-\b{q}}^T(z)\left(\hat{C}(\b{0},z)+\delta\hat{C}(\b{q},z)\right)\Phi_\b{q}(z)+\c{O}(\phi^3).
\end{align}
We define a 9-component vector of fluctuations,
\begin{align}\label{9vec}
\Phi_\b{q}(z)=\left(
\begin{array}{ccccccccc}
\tilde{\Delta}^{(+)} & \tilde{M}_2^{(+)} & \tilde{\Omega}_1^{(+)} & \tilde{M}_3^{(+)} & \tilde{\Omega}_2^{(+)}
& \tilde{M}_4^{(+)} & \tilde{\Omega}_3^{(+)} & \tilde{M}_1^{(+)} & \tilde{M}_5^{(+)}
\end{array}\right)_{\b{G},z}^T.
\end{align}
The $9\times 9$ matrix $\hat{C}(\b{0},z)$ contains all the $\hat{\Pi}^{(+)}_{J;m_J}$ operators of Sec.~\ref{sec:q0}:
\begin{align}\label{C0z}
\hat{C}(\b{0},z)=\left(\begin{array}{cccccc}
\hat{\Pi}_{0,2;0}^{(+)} & & & & & \\
& \hat{\Pi}_{1,2;x}^{(+)} & & & & \\
& & \hat{\Pi}_{1,2;y}^{(+)} & & & \\
& & & \hat{\Pi}_{1;0}^{(+)} & & \\
& & & & \hat{\Pi}_{2;\pm 2}^{(+)} & \\
& & & & & \hat{\Pi}_{2;\pm 2}^{(+)}
\end{array}\right).
\end{align}
Finally, $\delta\hat{C}(\b{q},z)$ contains the $\b{q}$-dependent pieces of the quadratic fluctuation kernel. It is quadratic in $\b{q}$, since the original GL functional (\ref{GL}) is quadratic in spatial derivatives. One can check that it is also explicitly invariant under $SO(2)_{J_z}$ rotations, which act simultaneously on the fields (\ref{9vec}) and on the wavevector $\b{q}$. Explicitly, the nonzero upper-triangular elements are given by:
\begin{align}\label{dCqz}
    \delta\hat{C}_{11}(\b{q},z)&=\delta\hat{C}_{44}(\b{q},z)=\delta\hat{C}_{66}(\b{q},z)=2K_1\b{q}^2,\nonumber\\
    \delta\hat{C}_{22}(\b{q},z)&=\delta\hat{C}_{77}(\b{q},z)=\delta\hat{C}_{88}(\b{q},z)=\delta\hat{C}_{99}(\b{q},z)=(2K_1+K_{23})\b{q}^2,\nonumber\\
    \delta\hat{C}_{33}(\b{q},z)&=2(K_1\b{q}^2+K_{23}q_y^2),\nonumber\\
    \delta\hat{C}_{55}(\b{q},z)&=2(K_1\b{q}^2+K_{23}q_x^2),\nonumber\\
    \delta\hat{C}_{13}(\b{q},z)&=-\sqrt{2}\delta\hat{C}_{24}(\b{q},z)=\sqrt{2}\delta\hat{C}_{48}(\b{q},z)
    =-\sqrt{2}\delta\hat{C}_{67}(\b{q},z)=-\sqrt{2}\delta\hat{C}_{69}(\b{q},z)=2iK_{23}q_y\partial_z,\nonumber\\
    -\delta\hat{C}_{15}(\b{q},z)&=-\sqrt{2}\delta\hat{C}_{26}(\b{q},z)=\sqrt{2}\delta\hat{C}_{47}(\b{q},z)
    =-\sqrt{2}\delta\hat{C}_{49}(\b{q},z)=-\sqrt{2}\delta\hat{C}_{68}(\b{q},z)=2iK_{23}q_x\partial_z,\nonumber\\
    \delta\hat{C}_{28}(\b{q},z)&=-\delta\hat{C}_{79}(\b{q},z)=K_{23}(q_x^2-q_y^2),\nonumber\\
    \delta\hat{C}_{29}(\b{q},z)&=-\delta\hat{C}_{35}(\b{q},z)=\delta\hat{C}_{78}(\b{q},z)=2K_{23}q_xq_y,
\end{align}
and the lower-triangular elements are given by $\delta\hat{C}_{ji}(\b{q},z)=\delta\hat{C}_{ij}(\b{q},z)$.

\subsection{``\texorpdfstring{$\b{k}\cdot\b{p}$}{kp}'' approach to computing the normal modes}\label{sec:kdotp}

We first review the basic idea of the $\b{k}\cdot\b{p}$ method in electronic bandstructure calculations~\cite{Winkler}, then explain how it can be applied to the diagonalization of the quadratic functional (\ref{FreeEnergyQuad}). This method constructs an effective Bloch Hamiltonian $H_{nn'}(\b{k})$ in the following way. The original Hamiltonian is
\begin{align}
\hat{H}=-\frac{\nabla^2}{2m}+V(\b{r}),
\end{align}
where $V(\b{r})$ is a periodic potential. Using Bloch's theorem, a given eigenstate is of the form $\psi_{\b{k}}(\b{r})=e^{i\b{k}\cdot\b{r}}u_{\b{k}}(\b{r})$ with eigenenergy $E(\b{k})$, where the periodic part of the wavefunction, $u_{\b{k}}(\b{r})$, obeys a $\b{k}$-dependent, reduced Schr\"odinger equation:
\begin{align}\label{redSchrod}
\left(-\frac{\nabla^2}{2m}+V(\b{r})+\frac{k^2}{2m}-\frac{i}{m}\b{k}\cdot\nabla\right)u_{\b{k}}(\b{r})=E(\b{k})u_{\b{k}}(\b{r}).
\end{align}
If we are interested in finding the bandstructure and wavefunctions in the vicinity of $\b{k}=0$, the idea of the $\b{k}\cdot\b{p}$ method is to first solve the $\b{k}=0$ problem,
\begin{align}\label{BandEdgeProblem}
\left(-\frac{\nabla^2}{2m}+V(\b{r})\right)u_{n\b{0}}(\b{r})=E_n(\b{0})u_{n\b{0}}(\b{r}),
\end{align}
which gives us a complete orthonormal set of states $|n\b{0}\rangle$, $u_{n\b{0}}(\b{r})=\langle\b{r}|n\b{0}\rangle$ (the band-edge wavefunctions), with discrete label $n$, since the equation is solved inside a single unit cell with periodic boundary conditions. We use this complete set as a basis to formally expand the unknown solution $u_\b{k}$ at $\b{k}\neq 0$:
\begin{align}\label{FormalExpanduk}
u_\b{k}(\b{r})=\sum_{n'} c_{n'\b{k}}u_{n'\b{0}}(\b{r}).
\end{align}
Substituting this expansion into (\ref{redSchrod}), multiplying on both sides by $u_{n\b{0}}^*(\b{r})$, and integrating over all $\b{r}$, we find that (\ref{redSchrod}) is converted into a matrix eigenvalue problem,
\begin{align}\label{MatEig}
\sum_{n'} H_{nn'}(\b{k})c_{n'\b{k}}=E(\b{k})c_{n\b{k}},
\end{align}
where the effective Bloch Hamiltonian is:
\begin{align}\label{Blocheff}
H_{nn'}(\b{k})=\left(E_n(\b{0})+\frac{k^2}{2m}\right)\delta_{nn'}+\frac{\b{k}}{m}\cdot\langle n\b{0}|-i\nabla|n'\b{0}\rangle,
\end{align}
where we have also used (\ref{BandEdgeProblem}). Knowing the band-edge wavefunctions $u_{n\b{0}}(\b{r})$, we can compute the matrix elements $\langle n\b{0}|-i\nabla|n'\b{0}\rangle=\int d\b{r}\,u_{n\b{0}}^*(\b{r})(-i\nabla)u_{n'\b{0}}(\b{r})$ explicitly, then diagonalize the matrix (\ref{Blocheff}) to obtain the eigenvalues $E(\b{k})$ and eigenvectors $c_{n\b{k}}$, from which we can reconstruct the original Bloch wavefunction via (\ref{FormalExpanduk}). Since the basis $\{u_{n\b{0}}(\b{r})\}$ is really infinite, in practice one keeps only the states with the $N$ lowest band-edge eigenvalues $E_n(\b{0})$, which, at least for sufficiently small $\b{k}$, should give a good approximation to the bandstructure in that range of energies.

We follow essentially the same approach for our problem. The analog of $u_\b{k}(\b{r})$ is $\Phi_\b{q}(z)$, and the analog of the reduced Schr\"odinger equation (\ref{redSchrod}) is:
\begin{align}\label{redSchrodGL}
\left(\hat{C}(\b{0},z)+\delta\hat{C}(\b{q},z)\right)\Phi_\b{q}(z)=\lambda(\b{q})\Phi_\b{q}(z).
\end{align}
The $\b{q}=0$ ``band-edge'' problem is:
\begin{align}
\hat{C}(\b{0},z)\Phi_{n\b{0}}(z)=\lambda_n(\b{0})\Phi_{n\b{0}}(z),
\end{align}
which, given the block-diagonal structure of the $\b{q}=0$ kernel (\ref{C0z}), breaks up into 6 independent systems of (coupled or individual) ordinary differential equations to be solved inside the ``unit cell'' $0<z<D$ with the appropriate Dirichlet/Neumann boundary conditions. In practice, we solve these differential equations by working with a truncated basis of sine ($\propto\sin(n\pi z/D)$) and cosine ($\propto\cos(n\pi z/D)$) functions and diagonalizing the resulting finite-dimensional matrix with standard linear algebra techniques. Finally, in the expansion analogous to (\ref{FormalExpanduk}),
\begin{align}
\Phi_\b{q}(z)=\sum_{n'}c_{n'\b{q}}\Phi_{n'\b{0}}(z),
\end{align}
we also keep a finite number of states $n'$. Proceeding as in the $\b{k}\cdot\b{p}$ problem, we obtain the matrix equation analogous to (\ref{MatEig}):
\begin{align}
\sum_{n'}H_{nn'}(\b{q})c_{n'\b{q}}=\lambda(\b{q})c_{n\b{q}},
\end{align}
where
\begin{align}\label{HG}
H_{nn'}(\b{q})=\lambda_n(\b{0})\delta_{nn'}+\langle n\b{0}|\delta\hat{C}(\b{q},z)|n'\b{0}\rangle,
\end{align}
and we denote
\begin{align}
\langle n\b{0}|\delta\hat{C}(\b{q},z)|n'\b{0}\rangle\equiv\int_0^Ddz\,\Phi_{n\b{0}}^\dag(z)\delta\hat{C}(\b{q},z)\Phi_{n'\b{0}}(z).
\end{align}
Using (\ref{dCqz}) and the sine and cosine basis functions mentioned earlier, the integration over $z$ can be performed analytically, and the matrix elements expressed in terms of the expansion coefficients (obtained numerically) for the ``band-edge wavefunctions'' $\Phi_{n\b{0}}(z)$. Diagonalizing the ``$\b{k}\cdot\b{p}$'' Hamiltonian $H_{nn'}(\b{q})$, we obtain the expansion coefficients $c_{n\b{q}}^{(j)}$ for the eigensolutions $\Phi_\b{q}^{(j)}(z)$ of the Schr\"odinger-like equation (\ref{redSchrodGL}),
\begin{align}
\Phi_\b{q}^{(j)}(z)=\sum_n c_{n\b{q}}^{(j)}\Phi_{n\b{0}}(z),
\end{align}
with eigenvalues $\lambda^{(j)}(\b{q})$. A generic fluctuation $\Phi_\b{q}(z)$ appearing in the free energy (\ref{FreeEnergyQuad}) can be expanded on this basis of eigensolutions:
\begin{align}\label{PhiGexp}
\Phi_\b{q}(z)=\sum_j u^{(j)}_{\b{q}}\Phi_\b{q}^{(j)}(z).
\end{align}
Using $\Phi_{-\b{q}}(z)=\Phi_\b{q}^*(z)$ and the orthonormality of the eigensolutions for each $\b{q}$,
\begin{align}
\int_0^Ddz\,\Phi_\b{q}^{(j)}(z)^\dag\Phi_\b{q}^{(j')}(z)=\delta_{jj'},
\end{align}
the fluctuation free energy density is given by
\begin{align}
f=\sum_\b{q}\sum_j\lambda^{(j)}(\b{q})|u^{(j)}_{\b{q}}|^2+\mathcal{O}(\phi^3),
\end{align}
which is Eq.~(6) in the main text.

\section{Cubic and quartic terms in the PDW functional}

\subsection{Cubic term}\label{sec:cubic}

As discussed earlier, at the quadratic level there is no mixing between real and imaginary modes. The mode $j=j_*$ that goes soft at the transition is a real mode; thus only real modes are kept in the Landau theory for the crystallization transition. At the cubic level, there can be mixing between real and imaginary modes; however, near the crystallization transition ($r\approx 0$ and $|\b{q}|\approx Q$) the imaginary modes have a high free-energy cost and their equilibrium amplitudes are small compared to $u_\b{q}=u^{(j_*)}_{\b{q}}$. Therefore we can justify ignoring the weak cubic mixing between real and imaginary modes. Expanding the original GL functional (\ref{GL}) to cubic order in $\phi$, we have:
\begin{align}\label{f3}
f^{(3)}=8\sum_{abc}\int_0^Ddz\,V_{abc}(\Delta_\parallel,\Delta_\perp)\Phi_a\Phi_b\Phi_c,
\end{align}
with $\Phi$ the 9-component vector (\ref{9vec}), $a,b,c=1,\ldots,9$, and the $V_{abc}$ coefficients are linear functions of the equilibrium order parameters $\Delta_\parallel$ and $\Delta_\perp$. Passing to the Fourier domain, we obtain:
\begin{align}
f^{(3)}=8\sum_{\b{q}_1,\b{q}_2,\b{q}_3}\sum_{abc}\int_0^D dz\,V_{abc}(\Delta_\parallel,\Delta_\perp)\Phi_{\b{q}_1,a}(z)\Phi_{\b{q}_2,b}(z)\Phi_{\b{q}_3,c}(z)\delta_{\b{q}_1+\b{q}_2+\b{q}_3,0}.
\end{align}
Substituting in $\Phi_\b{q}(z)\approx u_\b{q}\Phi_\b{q}^{(j_*)}(z)$, i.e., keeping only the soft mode, we have
\begin{align}
f^{(3)}\approx-\sum_{\b{q}_1,\b{q}_2,\b{q}_3}w(\b{q}_1,\b{q}_2,\b{q}_3)u_{\b{q}_1}u_{\b{q}_2}u_{\b{q}_3}\delta_{\b{q}_1+\b{q}_2+\b{q}_3,0},
\end{align}
where we define
\begin{align}
w(\b{q}_1,\b{q}_2,\b{q}_3)\equiv-8\sum_{abc}\int_0^D dz\,V_{abc}(\Delta_\parallel,\Delta_\perp)\Phi_{\b{q}_1,a}^{(j_*)}(z)\Phi_{\b{q}_2,b}^{(j_*)}(z)\Phi_{\b{q}_3,c}^{(j_*)}(z).
\end{align}
In general, the cubic vertex $w(\b{q}_1,\b{q}_2,\b{q}_3)$ has a nontrivial momentum dependence, which in real space corresponds to a nonlocal interaction $f^{(3)}=-\int d^2r_\parallel\,d^2r_\parallel'\,d^2r_\parallel''\,w(\b{r}_\parallel,\b{r}_\parallel',\b{r}_\parallel'')u(\b{r}_\parallel)u(\b{r}_\parallel')u(\b{r}_\parallel'')$. Following Ref.~\cite{ChaikinLubensky}, we want a simpler theory with a local interaction, $f^{(3)}\approx-w\int d^2r_\parallel\,u^3(\b{r}_\parallel)$. This corresponds to neglecting the momentum dependence of the cubic vertex $w(\b{q}_1,\b{q}_2,\b{q}_3)$. Since the ``wavefunction'' $\Phi_{\b{q}}^{(j_*)}$ is a smooth function of $\b{q}$, we approximate the cubic vertex by its value at zero momentum $\b{q}_1=\b{q}_2=\b{q}_3=0$, which can be viewed as the zeroth-order term in a gradient expansion:
\begin{align}
f^{(3)}\approx-w\sum_{\b{q}_1,\b{q}_2,\b{q}_3}u_{\b{q}_1}u_{\b{q}_2}u_{\b{q}_3}\delta_{\b{q}_1+\b{q}_2+\b{q}_3,0},
\end{align}
which is Eq.~(8) in the main text, where
\begin{align}\label{CubicVertexApprox}
w\equiv w(0,0,0)=-8\sum_{abc}\int_0^D dz\,V_{abc}(\Delta_\parallel,\Delta_\perp)\Phi_{\b{0},a}^{(j_*)}(z)\Phi_{\b{0},b}^{(j_*)}(z)\Phi_{\b{0},c}^{(j_*)}(z).
\end{align}
Since for $\b{q}=0$, $\Phi_\b{0}^{(j_*)}(z)$ is a ``band-edge wavefunction'' with well-defined rotational quantum numbers that belongs to the $\{\tilde{\Delta}^{(+)},\tilde{M}_2^{(+)}\}$ multiplet, only the terms in Eq.~(\ref{CubicVertexApprox}) with $a,b,c$ either 1 or 2 contribute. Explicitly, the relevant $V_{abc}$ coefficients are given by:
\begin{align}
V_{111}=\sqrt{2}\beta_{12345}\Delta_\perp,\hspace{5mm}
V_{112}=2\beta_{12}\Delta_\parallel,\hspace{5mm}
V_{122}=\sqrt{2}\beta_{12}\Delta_\perp,\hspace{5mm}
V_{222}=(2\beta_{12}+\beta_{345})\Delta_\parallel.
\end{align}

We can now understand why $w=0$ at $T_2^*$ but not at $T_1^*$. $T_2^*$ occurs in the planar phase, where $\Delta_\perp=0$ and thus $V_{111}=V_{122}=0$. The critical mode fluctuation in this case is a pure fluctuation of $\Delta_\perp$ about zero, i.e., $\tilde{\Delta}^{(+)}\neq 0$ but $\tilde{M}_2^{(+)}=0$. In other words, only the first component of $\Phi_{\b{0}}^{(j_*)}$ is nonzero, thus (\ref{CubicVertexApprox}) vanishes identically. In the PDB phase with $\Delta_\perp\neq 0$ and $\Delta_\parallel\neq 0$, the critical mode fluctuation has both $\tilde{\Delta}^{(+)}$ and $\tilde{M}_2^{(+)}$ components and $w$ is therefore generically nonzero. However, we must keep in mind that the planar phase is not realized in real $^3$He, thus the $T_2^*$ instability is not physical once strong-coupling corrections are taken into account. The study of the fluctuation spectrum in the A phase with strong-coupling corrections is left for future work.

\subsection{Quartic term}

A quartic term in the PDW functional is also important to ensure thermodynamic stability. This quartic term must have a positive coefficient, otherwise the free energy is unbounded from below and we have to go to higher orders in the Landau expansion. Going back to the original expansion (\ref{OPexpansion}), we see that since the original GL functional (\ref{GL}) is quartic in $A$, the leading contribution to the term quartic in fluctuations $\phi$ is simply obtained by evaluating the quartic term of (\ref{GL}) with $A=\phi$. In the Fourier domain, we have
\begin{align}
f^{(4)}=\sum_{\b{q}_1,\ldots,\b{q}_4}\sum_{abcd}V_{abcd}\int_0^Ddz\,\Phi_{\b{q}_1,a}(z)\Phi_{\b{q}_2,b}(z)
\Phi_{\b{q}_3,c}(z)\Phi_{\b{q}_4,d}(z)\delta_{\b{q}_1+\ldots+\b{q}_4,0},
\end{align}
where $V_{abcd}$ is now constant and only depends on the $\beta_i$ parameters. As for the cubic term, this would in general produce a momentum-dependent vertex $\eta(\b{q}_1,\ldots,\b{q}_4)$. Demanding a local quartic interaction $f^{(4)}\approx\eta\int d^2r_\parallel\,u^4(\b{r}_\parallel)$, we approximate the quartic vertex by its value at $\b{q}_i=0$, $i=1,\ldots,4$:
\begin{align}\label{QuarticVertex}
\eta=\sum_{abcd}V_{abcd}\int_0^Ddz\,\Phi_{\b{0},a}^{(j_*)}(z)\Phi_{\b{0},b}^{(j_*)}(z)\Phi_{\b{0},c}^{(j_*)}(z)\Phi_{\b{0},d}^{(j_*)}(z).
\end{align}
As for the cubic term, $\Phi_{\b{0}}^{(j_*)}(z)$ only has $\tilde{\Delta}^{(+)}$ and $\tilde{M}_2^{(+)}$ components, thus it is sufficient to substitute
\begin{align}
\phi=\Delta^{(+)}\lambda^0+M_2^{(+)}\lambda^2=\frac{1}{\sqrt{3}}\left(\tilde{\Delta}^{(+)}+\sqrt{2}\tilde{M}_2^{(+)}\right)\lambda^0+\frac{1}{\sqrt{3}}\left(\tilde{M}_2^{(+)}-\sqrt{2}\tilde{\Delta}^{(+)}\right)\lambda^2
=\left(\begin{array}{ccc}
\tilde{M}_2^{(+)} & & \\
& \tilde{M}_2^{(+)} & \\
& & \sqrt{2}\tilde{\Delta}^{(+)}
\end{array}\right)
\end{align}
in the quartic term of (\ref{GL}). Note that $\phi=\phi^*$ since only real modes participate (in the approximation where we neglect quartic mixing with high-energy imaginary modes, see Sec.~\ref{sec:cubic}), and also $\phi=\phi^T$, which together imply $\phi=\phi^\dag$. Substituting in (\ref{GL}), we obtain:
\begin{align}
F^{(4)}&=\int d^3r\left[\beta_{12}\left(\mathop{\mathrm{tr}}\phi^2\right)^2+\beta_{345}\mathop{\mathrm{tr}}\phi^4\right]\nonumber\\
&=\int d^3r\left[4\beta_{12345}\left(\tilde{\Delta}^{(+)}\right)^4+(4\beta_{12}+2\beta_{345})\left(\tilde{M}_2^{(+)}\right)^4+8\beta_{12}\left(\tilde{\Delta}^{(+)}\right)^2\left(\tilde{M}_2^{(+)}\right)^2\right],
\end{align}
thus the only relevant $V_{abcd}$ coefficients in Eq.~(\ref{QuarticVertex}) are
\begin{align}
V_{1111}=4\beta_{12345},\hspace{5mm}
V_{2222}=4\beta_{12}+2\beta_{345},\hspace{5mm}
V_{1122}=8\beta_{12},
\end{align}
which are all positive. Explicitly, we have:
\begin{align}
f^{(4)}\approx\eta\sum_{\b{q}_1,\ldots,\b{q}_4}u_{\b{q}_1}u_{\b{q}_2}u_{\b{q}_3}u_{\b{q}_4}\delta_{\b{q}_1+\ldots+\b{q}_4,0},
\end{align}
which is Eq.~(9) in the main text, with
\begin{align}
\eta=\int_0^Ddz\left[4\beta_{12345}\left(\Phi_{\b{0},1}^{(j_*)}(z)\right)^4
+(4\beta_{12}+2\beta_{345})\left(\Phi_{\b{0},2}^{(j_*)}(z)\right)^4
+8\beta_{12}\left(\Phi_{\b{0},1}^{(j_*)}(z)\right)^2\left(\Phi_{\b{0},2}^{(j_*)}(z)\right)^2\right]>0.
\end{align}
Thus, the PDW free energy functional is stable at quartic order.

\section{Symmetry analysis of the PDW order parameter}\label{sec:symmetry}

In the PDW phase, we set $u_\b{q}=u$ for $\b{q}\in\{\pm\b{G}_1,\pm\b{G}_2,\pm\b{G}_3\}$ and zero otherwise, where we can take $\b{G}_1=Q\hat{x}$, $\b{G}_2=Q(-\frac{1}{2}\hat{x}+\frac{\sqrt{3}}{2}\hat{y})$, and $\b{G}_3=-\b{G}_1-\b{G}_2=Q(-\frac{1}{2}\hat{x}-\frac{\sqrt{3}}{2}\hat{y})$. The superfluid order parameter thus becomes
\begin{align}
A_{\mu j}(\b{r})\approx\overline{A}_{\mu j}(z)+u\sum_{\b{q}=\pm\b{G}_1,\pm\b{G}_2,\pm\b{G}_3}\left(\Phi^{(j_*)}_\b{q}(z)\right)_{\mu j}e^{i\b{q}\cdot\b{r}_\parallel}.
\end{align}
It becomes more practical at this stage to define a new set of real $3\times 3$ matrices $\gamma^1,\ldots,\gamma^9$ as the linear combinations of Gell-Mann matrices (\ref{GellMann}) appropriate to the 9 components (\ref{9vec}) of the superfluid order parameter:
\begin{align}\label{gamma}
\gamma^1&=\frac{1}{\sqrt{3}}(\lambda^0-\sqrt{2}\lambda^2)=\left(\begin{array}{ccc}
0 & & \\
& 0 & \\
& & \sqrt{2}
\end{array}\right),\hspace{5mm}
\gamma^2=\frac{1}{\sqrt{3}}(\lambda^2+\sqrt{2}\lambda^0)=\left(\begin{array}{ccc}
1 & & \\
& 1 & \\
& & 0
\end{array}\right),\hspace{5mm}
\gamma^3=\frac{1}{\sqrt{2}}(i\lambda^6-\lambda^3)=\left(\begin{array}{ccc}
0 & & \\
& 0 & \\
& -\sqrt{2} & 0
\end{array}\right),\nonumber\\
\gamma^4&=\frac{1}{\sqrt{2}}(\lambda^3+i\lambda^6)=\left(\begin{array}{ccc}
0 & & \\
& 0 & \sqrt{2} \\
& & 0
\end{array}\right),\hspace{5mm}
\gamma^5=\frac{1}{\sqrt{2}}(i\lambda^7+\lambda^4)=\left(\begin{array}{ccc}
& & 0 \\
& 0 & \\
\sqrt{2} & &
\end{array}\right),\hspace{5mm}
\gamma^6=\frac{1}{\sqrt{2}}(\lambda^4-i\lambda^7)=\left(\begin{array}{ccc}
& & \sqrt{2} \\
& 0 & \\
0 & &
\end{array}\right),\nonumber\\
\gamma^7&=i\lambda^8=\left(\begin{array}{ccc}
0 & 1 & \\
-1 & 0 & \\
& & 0
\end{array}\right),\hspace{5mm}
\gamma^8=\lambda^1=\left(\begin{array}{ccc}
1 & & \\
& -1 & \\
& & 0
\end{array}\right),\hspace{5mm}
\gamma^9=\lambda^5=\left(\begin{array}{ccc}
0 & 1 & \\
1 & 0 & \\
& & 0
\end{array}\right).
\end{align}
Focusing on the pure PDW component $\phi_{\mu j}\equiv A_{\mu j}-\overline{A}_{\mu j}$, we have
\begin{align}\label{PDWOP}
\phi_{\mu j}(\b{r})=u\sum_\alpha\gamma_{\mu j}^\alpha\sum_{i=1}^3\left(\Phi^{(j_*)}_{\b{G}_i}(z)_\alpha e^{i\b{G}_i\cdot\b{r}_\parallel}+\Phi_{-\b{G}_i}^{(j_*)}(z)_\alpha e^{-i\b{G}_i\cdot\b{r}_\parallel}\right),
\end{align}
where $\Phi^{(j_*)}_\b{q}(z)_1,\ldots,\Phi^{(j_*)}_\b{q}(z)_9$ are the 9 components of (\ref{9vec}).

In practice, we calculate $\Phi^{(j_*)}_\b{q}(z)$ numerically for a single wavevector $\b{q}=Q\hat{x}=\b{G}_1$, and obtain the other $\Phi^{(j_*)}_{\b{G}_i}(z)$ and $\Phi_{-\b{G}_i}^{(j_*)}(z)$ by symmetry. First, $\Phi_{\pm\b{G}_2}^{(j_*)}(z)$ and $\Phi_{\pm\b{G}_3}^{(j_*)}(z)$ can be obtained from $\Phi_{\pm\b{G}_1}^{(j_*)}(z)$ by $C_6$ rotations. $\hat{C}(\b{q},z)$ possesses a continuous $SO(2)_{J_z}$ rotation symmetry, expressed as
\begin{align}\label{RdagCR}
\c{R}^{-1}(\theta)\hat{C}(R_\theta\b{q},z)\c{R}(\theta)=\hat{C}(\b{q},z),
\end{align}
where
\begin{align}
R_\theta=\left(\begin{array}{cc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{array}\right),
\end{align}
is the ordinary $SO(2)$ rotation matrix, and $\c{R}(\theta)=e^{-i\theta\c{J}_z}$ is a 9-dimensional representation of $SO(2)$ with generator
\begin{align}\label{curlyJz}
\c{J}_z=\left(\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right)\oplus
\left(\begin{array}{cccc}
& & -i & \\
& & & i \\
i & & & \\
& -i & &
\end{array}\right)\oplus
\left(\begin{array}{c}
0
\end{array}\right)\oplus
\left(\begin{array}{cc}
& -2i \\
2i &
\end{array}\right).
\end{align}
Explicitly, we have
\begin{align}\label{curlyR}
\c{R}(\theta)=\left(\begin{array}{cc}
1 &  \\
 & 1
\end{array}\right)\oplus
\left(\begin{array}{cccc}
\cos\theta & 0 & -\sin\theta & 0 \\
0 & \cos\theta & 0 & \sin\theta \\
\sin\theta & 0 & \cos\theta & 0 \\
0 & -\sin\theta & 0 & \cos\theta
\end{array}\right)\oplus
\left(\begin{array}{c}
1
\end{array}\right)\oplus
\left(\begin{array}{cc}
\cos 2\theta & -\sin 2\theta \\
\sin 2\theta & \cos 2\theta
\end{array}\right).
\end{align}
Equation (\ref{RdagCR}) implies that if $\Phi_\b{q}^{(j)}(z)$ is an eigenstate of $\hat{C}(\b{q},z)$ with eigenvalue $\lambda^{(j)}(\b{q})$, then $\c{R}(\theta)\Phi_\b{q}^{(j)}(z)$ is an eigenstate of $\hat{C}(R_\theta\b{q},z)$ with the same eigenvalue. Since $\b{G}_2=R_{2\pi/3}\b{G}_1$ and $\b{G}_3=R_{4\pi/3}\b{G}_1$, we can thus take $\Phi^{(j_*)}_{\b{G}_i}(z)=\c{R}(\theta_i)\Phi^{(j_*)}_{\b{G}_1}(z)$, $i=1,2,3$, with $\theta_1=0$, $\theta_2=2\pi/3$, and $\theta_3=4\pi/3$. Likewise, $\Phi^{(j_*)}_{-\b{G}_i}(z)=\c{R}(\theta_i)\Phi^{(j_*)}_{-\b{G}_1}(z)$. Equation (\ref{PDWOP}) thus becomes
\begin{align}\label{PDWOP2}
\phi_{\mu j}(\b{r})=u\sum_\alpha\gamma_{\mu j}^\alpha\sum_{i=1}^3\c{R}_{\alpha\beta}(\theta_i)\left(\Phi_{\b{G}_1}^{(j_*)}(z)_\beta e^{i\b{G}_i\cdot\b{r}_\parallel}+\Phi_{-\b{G}_1}^{(j_*)}(z)_\beta e^{-i\b{G}_i\cdot\b{r}_\parallel}\right).
\end{align}
Next, $\Phi^{(j_*)}_{-\b{G}_1}(z)$ is related to $\Phi^{(j_*)}_{\b{G}_1}(z)$ by a reflection symmetry $x\rightarrow-x$. One can check explicitly that
\begin{align}\label{Mx}
\c{M}_x^{-1}\hat{C}(-q_x,q_y;z)\c{M}_x=\hat{C}(q_x,q_y;z),
\end{align}
where $\c{M}_x=\mathrm{diag}(1,1,1,1,-1,-1,-1,1,-1)$. In particular, $\c{M}_x^{-1}\hat{C}(-\b{G}_1,z)\c{M}_x=\hat{C}(\b{G}_1,z)$, thus we can choose $\Phi_{-\b{G}_1}^{(j_*)}(z)=\c{M}_x\Phi_{\b{G}_1}^{(j_*)}(z)$. Since $\c{M}_x$ is diagonal, Eq.~(\ref{PDWOP2}) becomes
\begin{align}\label{PDWOP3}
\phi_{\mu j}(\b{r})=u\sum_{\alpha\beta}\gamma_{\mu j}^\alpha\sum_{i=1}^3\c{R}_{\alpha\beta}(\theta_i)
\left(e^{i\b{G}_i\cdot\b{r}_\parallel}+\c{M}_x^{\beta\beta}e^{-i\b{G}_i\cdot\b{r}_\parallel}\right)
\Phi_{\b{G}_1}^{(j_*)}(z)_\beta.
\end{align}
Since $\c{M}_x^{\beta\beta}=\pm 1$, the $\b{r}_\parallel$ dependence of the PDW order parameter will be via linear combinations of $\cos\b{G}_i\cdot\b{r}_\parallel$ and $\sin\b{G}_i\cdot\b{r}_\parallel$.

Further simplifications occur due to the fact that $\hat{C}(\b{G}_1,z)$ should be invariant under a $y\rightarrow-y$ symmetry, since $\b{G}_1$ is entirely along the $x$ axis. Analogously to Eq.~(\ref{Mx}), one can check that
\begin{align}
\c{M}_y^{-1}\hat{C}(q_x,-q_y;z)\c{M}_y=\hat{C}(q_x,q_y;z),
\end{align}
where $\c{M}_y=\mathrm{diag}(1,1,-1,-1,1,1,-1,1,-1)$. In particular, $[\c{M}_y,\hat{C}(\b{G}_1,z)]=0$, thus $\Phi^{(j_*)}_{\b{G}_1}(z)$ must be an eigenvector of $\c{M}_y$ with eigenvalue $\pm 1$, since $\c{M}_y^2=1$. Since $\c{M}_y$ is diagonal, we can easily write down its action on the 9 components of $\Phi^{(j_*)}_{\b{G}_1}(z)$. Denoting $\Phi\equiv\Phi^{(j_*)}_{\b{G}_1}(z)$ for simplicity, we have
\begin{align}\label{My}
(\c{M}_y\Phi)_{1,2,5,6,8}=\Phi_{1,2,5,6,8},\hspace{5mm}
(\c{M}_y\Phi)_{3,4,7,9}=-\Phi_{3,4,7,9}.
\end{align}
There are two possibilities. If $\c{M}_y\Phi=\Phi$, then $\Phi_{3,4,7,9}$ must all vanish. Conversely, if $\c{M}_y\Phi=-\Phi$, then $\Phi_{1,2,5,6,8}$ must all vanish. We find numerically that near the PDW instability, the $\c{M}_y$ eigenvalue of $\Phi$ is $+1$, thus $\Phi_{3,4,7,9}=0$.

\subsection{Structure of the PDW order parameter: group-theoretic analysis}

By assuming $u_\b{q}=u$ for all $\b{q}\in\{\pm\b{G}_1,\pm\b{G}_2,\pm\b{G}_3\}$, we have constructed an order parameter $\phi$ that is invariant under the action of a spin-orbital point group $D_6^{L_z+S_z}$ which acts simultaneously on spin and orbital coordinates:
\begin{align}\label{D6Jzaction}
    \c{D}(g)\phi(g^{-1}\b{r}_\parallel,z)\c{D}^{-1}(g)=\phi(\b{r}_\parallel,z),
\end{align}
where $g$ is any element of the point group $D_6$ of the triangular lattice. In the main text, we express $\phi$ in the form
\begin{align}
    \phi=u\sum_{j,k}\phi^{(j,k)}(z)X^{(j,k)}(\b{r}_\parallel),
\end{align}
where $X^{(j,k)}(\b{r}_\parallel)$ is the $k$th $D_6^{L_z+S_z}$ invariant---i.e., a $3\times 3$ matrix transforming according to Eq.~(\ref{D6Jzaction})---associated with the irreducible representation (irrep) $j$ of $D_6$, and $\phi^{(j,k)}(z)$ is the corresponding nonuniversal amplitude (a scalar function of $z$). We now explain how the $X$ invariants can be constructed~\cite{BirPikus} and give a precise definition of the amplitudes plotted in Fig.~4(a).

The dihedral group $D_6$ is a finite group of order 12 consisting of 6 rotations $C_6^0=e,C_6,\ldots,C_6^5$ and 6 reflections $\sigma_1,\sigma_2,\sigma_3,\sigma_1',\sigma_2',\sigma_3'$. We define $C_6$ as a counterclockwise rotation by $\pi/3$, $\sigma_1,\sigma_2,\sigma_3$ as the reflections that leave $\b{G}_1,\b{G}_2,\b{G}_3$ invariant, respectively, and $\sigma_1',\sigma_2',\sigma_3'$ as the reflections that map $\b{G}_1$ to $\b{G}_2$, $\b{G}_2$ to $\b{G}_3$, and $\b{G}_3$ to $-\b{G}_1$, respectively. $D_6$ has 2 generators, which we take to be $C_6$ and $\sigma_1$, and 6 conjugacy classes, which is also the number of irreps. The theory of invariants is based on the fact that the trivial representation only appears in the direct product of two mutually conjugate representations~\cite{BirPikus}; thus invariant functions can be formed only from the product of basis functions belonging to a representation and its conjugate. To enumerate all possible invariants, we need look only at all possible irreps of $D_6$, which are the $A_1$, $A_2$, $B_1$, $B_2$, $E_1$, and $E_2$ irreps. (These irreps are all real and thus self-conjugate.) In the present single-harmonic approximation, the PDW order parameter (\ref{PDWOP}) is expressed as a linear combination of 6 scalar $\b{r}_\parallel$-dependent functions
\begin{align}\label{psi}
    \psi_1=\cos\b{G}_1\cdot\b{r}_\parallel,\hspace{5mm}
    \psi_2=\sin\b{G}_1\cdot\b{r}_\parallel,\hspace{5mm}
    \psi_3=\cos\b{G}_2\cdot\b{r}_\parallel,\hspace{5mm}
    \psi_4=\sin\b{G}_2\cdot\b{r}_\parallel,\hspace{5mm}
    \psi_5=\cos\b{G}_3\cdot\b{r}_\parallel,\hspace{5mm}
    \psi_6=\sin\b{G}_3\cdot\b{r}_\parallel,\hspace{5mm}
\end{align}
multiplying 9 matrices $\gamma^1,\ldots,\gamma^9$, with $z$-dependent coefficients. The 6 functions (\ref{psi}) transform according to a (reducible) 6-dimensional representation of $D_6$, and the 9 matrices (\ref{gamma}) according to a (reducible) 9-dimensional representation. To construct the $X$ invariants, we must first decompose those reducible representations into a direct sum of irreps and construct irreducible basis functions that transform according to the latter. Then, we may form invariant products for each irrep.

We begin with the 6-dimensional representation $M$ according to which the $\psi_j(\b{r}_\parallel)$, $j=1,\ldots,6$ transform:
\begin{align}
    \psi_j(g^{-1}\b{r}_\parallel)=\sum_i\psi_i(\b{r}_\parallel)M_{ij}(g),\hspace{5mm}g\in D_6.
\end{align}
$M$ can be defined by its expression on the generators,
\begin{align}
    M(C_6)=\left(\begin{array}{cccccc}
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & -1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 0 & 0 & -1 \\
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 & 0 & 0
    \end{array}\right),\hspace{10mm}
    M(\sigma_1)=\left(\begin{array}{cccccc}
    1 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 \\
    0 & 0 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 0 & 0
    \end{array}\right).
\end{align}
We next need the representation matrices $D^{(j)}(g)$ for the 6 irreps $j=A_1,A_2,B_1,B_2,E_1,E_2$ of $D_6$. These are given by:
\begin{align}\label{Dirreps}
    D^{(A_1)}(C_6)&=1,\hspace{5mm}D^{(A_1)}(\sigma_1)=1;\nonumber\\
    D^{(A_2)}(C_6)&=1,\hspace{5mm}D^{(A_2)}(\sigma_1)=-1;\nonumber\\
    D^{(B_1)}(C_6)&=-1,\hspace{5mm}D^{(B_1)}(\sigma_1)=1;\nonumber\\
    D^{(B_2)}(C_6)&=-1,\hspace{5mm}D^{(B_2)}(\sigma_1)=-1;\nonumber\\
    D^{(E_1)}(C_6)&=\left(\begin{array}{cc}
    \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
    \frac{\sqrt{3}}{2} & \frac{1}{2}
    \end{array}\right),\hspace{5mm}
    D^{(E_1)}(\sigma_1)=\left(\begin{array}{cc}
    1 & 0 \\
    0 & -1
    \end{array}\right);\nonumber\\
    D^{(E_2)}(C_6)&=\left(\begin{array}{cc}
    -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\
    \frac{\sqrt{3}}{2} & -\frac{1}{2}
    \end{array}\right),\hspace{5mm}
    D^{(E_2)}(\sigma_1)=\left(\begin{array}{cc}
    1 & 0 \\
    0 & -1
    \end{array}\right).
\end{align}
The reducible representation $M$ decomposes as the direct sum $M=\bigoplus_j a_jD^{(j)}=A_1\oplus B_1\oplus E_1\oplus E_2$, where the $a_j$ are obtained from $a_j=\frac{1}{h}\sum_{g\in G}\chi^{(j)}(g)^*\chi^{(M)}(g)$, where $h=12$ is the order of the group $G=D_6$, $\chi^{(j)}$ is the character of irrep $j$, and $\chi^{(M)}$ is the character of $M$. To construct basis functions belonging to a given irrep $j$, we use the fact that for fixed $\nu=1,\ldots,d_j$ with $d_j=\dim D^{(j)}$, the $6\times 6$ matrices
\begin{align}
    \Pi_{\mu\nu}^{(j)}=\frac{d_j}{h}\sum_{g\in G}D_{\mu\nu}^{(j)}(g)^*M(g),\hspace{5mm}\mu=1,\ldots,d_j,
\end{align}
project a function onto irrep $j$. For simplicity, we set $\nu=1$ and consider the matrices
\begin{align}
    P_\mu^{(j)}=\frac{d_j}{h}\sum_{g\in G}D_{\mu 1}^{(j)}(g)^*M(g),\hspace{5mm}\mu=1,\ldots,d_j.
\end{align}
These matrices are projectors in the sense that for an arbitrary 6-component vector $\b{f}$, the vectors $P_\mu^{(j)}\b{f}$, $\mu=1,\ldots,d_j$ transform among themselves according to irrep $j$:
\begin{align}\label{projtrans}
    M(g)P_\nu^{(j)}\b{f}=\sum_\mu\left(P_\mu^{(j)}\b{f}\right)D_{\mu\nu}^{(j)}(g),\hspace{5mm}g\in D_6.
\end{align}
Using this, we easily construct the following unit vectors,
\begin{align}
    \hat{\b{e}}^{(A_1)}&=\frac{1}{\sqrt{3}}(1,0,1,0,1,0),\nonumber\\
    \hat{\b{e}}^{(B_1)}&=\frac{1}{\sqrt{3}}(0,1,0,1,0,1),\nonumber\\
    \hat{\b{e}}_1^{(E_1)}&=\frac{1}{\sqrt{6}}(0,2,0,-1,0,-1),\hspace{5mm}\hat{\b{e}}_2^{(E_1)}=\frac{1}{\sqrt{2}}(0,0,0,1,0,-1),\nonumber\\
    \hat{\b{e}}_1^{(E_2)}&=\frac{1}{\sqrt{6}}(2,0,-1,0,-1,0),\hspace{5mm}\hat{\b{e}}_2^{(E_2)}=\frac{1}{\sqrt{2}}(0,0,-1,0,1,0),
\end{align}
which transform as $M(g)\hat{\b{e}}_\nu^{(j)}=\sum_\mu\hat{\b{e}}_\mu^{(j)}D_{\mu\nu}^{(j)}(g)$ as per Eq.~(\ref{projtrans}), where the $\mu,\nu$ indices can be suppressed for one-dimensional irreps. Note that the $P_\mu^{(j)}$ vanish identically for $A_2$ and $B_2$ since those irreps do not appear in the direct-sum decomposition of $M$. Using those unit vectors, we can construct irreducible basis functions $\Psi_\mu^{(j)}(\b{r}_\parallel)=\hat{\b{e}}_\mu^{(j)}\cdot\boldsymbol{\psi}(\b{r}_\parallel)$, where $\boldsymbol{\psi}=(\psi_1,\ldots,\psi_6)$ is given in Eq.~(\ref{psi}). These irreducible basis functions transform as
\begin{align}
    \Psi_\nu^{(j)}(g^{-1}\b{r}_\parallel)=\sum_\mu\Psi_\mu^{(j)}(\b{r}_\parallel)D_{\mu\nu}^{(j)}(g),\hspace{5mm}g\in D_6.
\end{align}

We now repeat this exercise for the 9-dimensional representation $\Gamma$ according to which the matrices $\gamma^\alpha$, $\alpha=1,\ldots,9$ transform:
\begin{align}
    \c{D}(g)\gamma^\beta\c{D}^{-1}(g)=\sum_\alpha\gamma^\alpha\Gamma_{\alpha\beta}(g),\hspace{5mm}g\in D_6.
\end{align}
Here $\c{D}$ is itself a 3-dimensional representation of $D_6$ consisting of $O(3)\supset D_6$ matrices; its $SO(3)$ subgroup corresponds to spin-orbital rotations discussed in Eq.~(\ref{SOrotation}), thus $\c{D}(C_6)=e^{-i\pi\lambda^8/3}$. Since $\sigma_1$ is a reflection in the $y$ direction, we thus have:
\begin{align}
    \c{D}(C_6)=\left(\begin{array}{ccc}
    \frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
    \frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
    0 & 0 & 1
    \end{array}\right),\hspace{10mm}
    \c{D}(\sigma_1)=\left(\begin{array}{ccc}
    1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & 1
    \end{array}\right).
\end{align}
Comparing with Eq.~(\ref{Dirreps}), it is clear that $\c{D}=E_1\oplus A_1$, with the $A_1$ factor corresponding to the trivial action of $D_6$ on the $z$ coordinate. Turning to the $\Gamma$ representation, we have, in the notation of Eqs.~(\ref{curlyJz}-\ref{curlyR}), $\Gamma(C_6)=e^{-i\pi\c{J}_z/3}=\c{R}(\pi/3)$, and by explicit computation we obtain $\Gamma(\sigma_1)=\mathrm{diag}(1,1,-1,-1,1,1,-1,1,-1)=\c{M}_y$. This reducible representation decomposes as $\Gamma=\bigoplus_j a_jD^{(j)}=A_1\oplus A_1\oplus A_2\oplus E_1\oplus E_1\oplus E_2$, where the multiplicities are obtained from $a_j=\frac{1}{h}\sum_{g\in G}\chi^{(j)}(g)^*\chi^{(\Gamma)}(g)$. To construct irreducible basis functions (which in this case will be $3\times 3$ basis matrices), we again form the projectors
\begin{align}
    P_\mu^{(j)}=\frac{d_j}{h}\sum_{g\in G}D_{\mu 1}^{(j)}(g)^*\Gamma(g).
\end{align}
Because of the two-fold multiplicities of the $A_1$ and $E_1$ irreps inside $\Gamma$, we expect there will be two sets of $A_1$ and $E_1$ basis matrices. This expresses itself in the fact that for those irreps, the action of $P_\mu^{(j)}$ on a generic 9-component vector $\b{f}$ will give a linear combination of two linearly independent basis vectors. In this way we construct the following unit vectors:
\begin{align}\label{basisvecs9D}
    \hat{\b{e}}^{(A_1,1)}&=(1,0,0,0,0,0,0,0,0),\nonumber\\
    \hat{\b{e}}^{(A_1,2)}&=(0,1,0,0,0,0,0,0,0),\nonumber\\
    \hat{\b{e}}^{(A_2)}&=(0,0,0,0,0,0,1,0,0),\nonumber\\
    \hat{\b{e}}_1^{(E_1,1)}&=(0,0,0,0,1,0,0,0,0),\hspace{5mm}\hat{\b{e}}_2^{(E_1,1)}=(0,0,-1,0,0,0,0,0,0),\nonumber\\
    \hat{\b{e}}_1^{(E_1,2)}&=(0,0,0,0,0,1,0,0,0),\hspace{5mm}\hat{\b{e}}_2^{(E_1,2)}=(0,0,0,1,0,0,0,0,0),\nonumber\\
    \hat{\b{e}}_1^{(E_2)}&=(0,0,0,0,0,0,0,1,0),\hspace{5mm}\hat{\b{e}}_2^{(E_2)}=(0,0,0,0,0,0,0,0,1),
\end{align}
which transform as $\Gamma(g)\hat{\b{e}}_\nu^{(j)}=\sum_\mu\hat{\b{e}}_\mu^{(j)}D_{\mu\nu}^{(j)}(g)$, where the $\mu,\nu$ indices are again suppressed for one-dimensional irreps. Here $\hat{\b{e}}^{(A_1,1)}$ and $\hat{\b{e}}^{(A_1,2)}$ denote two independent basis vectors transforming in the $A_1$ irrep, and likewise $\hat{\b{e}}_\mu^{(E_1,1)}$ and $\hat{\b{e}}_\mu^{(E_1,2)}$ are two independent sets of basis vectors transforming in the $E_1$ irrep. Also note that $P_\mu^{(B_1)}=P_\mu^{(B_2)}=0$ since those irreps do not appear in the direct-sum decomposition of $\Gamma$. From the unit vectors (\ref{basisvecs9D}) and $\boldsymbol{\gamma}\equiv(\gamma^1,\ldots,\gamma^9)$ we form irreducible basis matrices $\Lambda_\mu^{(j)}=\hat{\b{e}}_\mu^{(j)}\cdot\boldsymbol{\gamma}$, which transform as
\begin{align}
    \c{D}(g)\Lambda_\nu^{(j)}\c{D}^{-1}(g)=\sum_\mu\Lambda_\mu^{(j)}D_{\mu\nu}^{(j)}(g),\hspace{5mm}g\in D_6.
\end{align}

From the irreducible basis functions $\Psi_\mu^{(j)}(\b{r}_\parallel)$ and irreducible basis matrices $\Lambda_\mu^{(j)}$, we may finally form $D_6$ invariants via
\begin{align}
    X^{(j,\alpha\beta)}(\b{r}_\parallel)=\sum_{\mu=1}^{d_j}\Lambda_\mu^{(j,\alpha)}\Psi_\mu^{(j,\beta)}(\b{r}_\parallel)^*,
\end{align}
where $\Lambda_\mu^{(j,\alpha)}$ is the $\alpha$th irreducible basis matrix belonging to irrep $j$, and $\Psi_\mu^{(j,\beta)}(\b{r}_\parallel)$ is the $\beta$th irreducible basis function belonging to irrep $j$. (Since the irreps and associated basis functions are real, we may in fact drop the complex conjugate.) We note that only the $A_1$, $E_1$, and $E_2$ irreps appear in both the direct-sum decompositions of $M$ and $\Gamma$; thus only for those irreps can invariants be constructed. Explicitly, the relevant irreducible basis functions are thus:
\begin{align}\label{irrbasisfcts}
    \Psi^{(A_1)}(\b{r}_\parallel)&=\hat{\b{e}}^{(A_1)}\cdot\boldsymbol{\psi}=\frac{1}{\sqrt{3}}\left(\cos Qx+2\cos\textstyle\frac{1}{2}Qx\cos\frac{\sqrt{3}}{2}Qy\right),\nonumber\\
    \boldsymbol{\Psi}^{(E_1)}(\b{r}_\parallel)&=\left(\hat{\b{e}}_1^{(E_1)}\cdot\boldsymbol{\psi},\hat{\b{e}}_2^{(E_1)}\cdot\boldsymbol{\psi}\right)
    =\sqrt{\frac{2}{3}}\left(\sin Qx+\sin\textstyle\frac{1}{2}Qx\cos\frac{\sqrt{3}}{2}Qy,\sqrt{3}\cos\frac{1}{2}Qx\sin\frac{\sqrt{3}}{2}Qy\right),\nonumber\\
    \boldsymbol{\Psi}^{(E_2)}(\b{r}_\parallel)&=\left(\hat{\b{e}}_1^{(E_2)}\cdot\boldsymbol{\psi},\hat{\b{e}}_2^{(E_2)}\cdot\boldsymbol{\psi}\right)
    =\sqrt{\frac{2}{3}}\left(\cos Qx-\cos\textstyle\frac{1}{2}Qx\cos\frac{\sqrt{3}}{2}Qy,-\sqrt{3}\sin\frac{1}{2}Qx\sin\frac{\sqrt{3}}{2}Qy\right),
\end{align}
which can be understood as $s$-wave, $p$-wave, and $d$-wave triangular lattice harmonics, respectively (see, e.g., Ref.~\cite{zhou2008}, where slightly different linear combinations are chosen). Likewise, the $A_1,E_1,E_2$ irreducible basis matrices are:
\begin{align}\label{irrbasismat}
    \Lambda^{(A_1,1)}&=\hat{\b{e}}^{(A_1,1)}\cdot\boldsymbol{\gamma}=\gamma^1,\nonumber\\
    \Lambda^{(A_1,2)}&=\hat{\b{e}}^{(A_1,2)}\cdot\boldsymbol{\gamma}=\gamma^2,\nonumber\\
    \boldsymbol{\Lambda}^{(E_1,1)}&
    =\left(\hat{\b{e}}_1^{(E_1,1)}\cdot\boldsymbol{\gamma},\hat{\b{e}}_2^{(E_1,1)}\cdot\boldsymbol{\gamma}\right)
    =(\gamma^5,-\gamma^3),\nonumber\\
    \boldsymbol{\Lambda}^{(E_1,2)}&
    =\left(\hat{\b{e}}_1^{(E_1,2)}\cdot\boldsymbol{\gamma},\hat{\b{e}}_2^{(E_1,2)}\cdot\boldsymbol{\gamma}\right)
    =(\gamma^6,\gamma^4),\nonumber\\
    \boldsymbol{\Lambda}^{(E_2)}&
    =\left(\hat{\b{e}}_1^{(E_2)}\cdot\boldsymbol{\gamma},\hat{\b{e}}_2^{(E_2)}\cdot\boldsymbol{\gamma}\right)
    =(\gamma^8,\gamma^9).
\end{align}
Therefore, only 5 invariants can be constructed: $\Lambda^{(A_1,1)}\Psi^{(A_1)}$, $\Lambda^{(A_1,2)}\Psi^{(A_1)}$, $\b{\Lambda}^{(E_1,1)}\cdot\b{\Psi}^{(E_1)}$, $\b{\Lambda}^{(E_1,2)}\cdot\b{\Psi}^{(E_1)}$, and $\b{\Lambda}^{(E_2)}\cdot\b{\Psi}^{(E_2)}$. In fact, using the explicit form of the matrices $\c{R}(\theta_i)$ and $\c{M}_x$, we can show that the PDW order parameter (\ref{PDWOP3}) is a linear combination of those 5 invariants:
\begin{align}\label{PDWOP4}
\phi_{\mu j}(\b{r})&=2\sqrt{3}u\left(\Phi_1(z)\Lambda^{(A_1,1)}_{\mu j}+\Phi_2(z)\Lambda^{(A_1,2)}_{\mu j}\right)\Psi^{(A_1)}(\b{r}_\parallel)\nonumber\\
&\phantom{=}+i\sqrt{6}u\left(\Phi_5(z)\b{\Lambda}^{(E_1,1)}_{\mu j}+\Phi_6(z)\b{\Lambda}^{(E_1,2)}_{\mu j}\right)\cdot\b{\Psi}^{(E_1)}(\b{r}_\parallel)\nonumber\\
&\phantom{=}+\sqrt{6}u\Phi_8(z)\b{\Lambda}^{(E_2)}_{\mu j}\cdot\b{\Psi}^{(E_2)}(\b{r}_\parallel),
\end{align}
where we recall the notation $\Phi_\beta(z)\equiv\Phi_{\b{G}_1}^{(j_*)}(z)_\beta$ employed earlier, see Eq.~(\ref{My}).

Finally, we discuss the consequences of time-reversal symmetry. The ``Hamiltonian'' $\hat{C}(\b{q},z)$ has the property $\hat{C}(\b{q},z)^*=\hat{C}(-\b{q},z)$ as can be checked explicitly. This implies the spectrum of $\hat{C}$ is symmetric under $\b{q}\rightarrow-\b{q}$. Since $\Phi_{\b{G}_1}^{(j_*)}(z)$ is nondegenerate in the region of interest (near and at the instability), we can thus choose $\Phi_{-\b{G}_1}^{(j_*)}(z)=\Phi_{\b{G}_1}^{(j_*)}(z)^*$. However, we saw earlier that reflection symmetry in the $x$ direction implies that we can choose $\Phi_{-\b{G}_1}^{(j_*)}(z)=\c{M}_x\Phi_{\b{G}_1}^{(j_*)}(z)$. Combining these two equalities, and using the explicit diagonal form of $\c{M}_x$ presented earlier, we find that $\Phi_{1,2,3,4,8}$ can be made real (in a suitable choice of gauge), while $\Phi_{5,6,7,9}$ is purely imaginary. Together with the earlier numerical observation that $\Phi_{3,4,7,9}=0$, we find that $\Phi_{1,2,8}$ is real and $\Phi_{5,6}$ purely imaginary, which can be confirmed numerically. Since the basis functions (\ref{irrbasisfcts}) and basis matrices (\ref{irrbasismat}) are real, this implies that the PDW order parameter (\ref{PDWOP4}) is real for real $u$, as expected. The real amplitudes $\Phi_1(z)$, $\Phi_2(z)$, $i\Phi_5(z)$, $i\Phi_6(z)$, and $\Phi_8(z)$ are plotted in Fig.~4(a) of the main text and also reproduced here in Fig.~\ref{fig:zamps}. To aid in the visualization of the three-dimensional structure of the PDW, we also give animated plots of the nine components $\phi_{11}(\b{r})$, $\phi_{12}(\b{r})$, $\ldots$, $\phi_{33}(\b{r})$ of the PDW order parameter (\ref{PDWOP4}) in units of $u$ in a set of nine ancillary files \texttt{phi11.gif}, \texttt{phi12.gif}, $\ldots$, \texttt{phi33.gif}.

\begin{figure}[t]
    \includegraphics[width=0.5\columnwidth]{Figure4aSI.pdf}
    \caption{Irreducible $z$-dependent amplitudes of the PDW order parameter.}
    \label{fig:zamps}
    \centering
\end{figure}

redpajama_set_name: RedPajamaArXiv