TITLE: Applications of cluster algebras
QUESTION [14 upvotes]: Why are so many algebraists nowadays interested in cluster algebras?
(This is a rewording of one half of the closed question Cluster algebras and teichmuller theory.)

REPLY [10 votes]: One reason is that cluster algebras have motivated many recent developments in the representation theory of associative algebras. There is a lot one can say about this, so I will try to just give an overview of some of the key ideas, and suggest further reading. I recommend Keller's survey article (https://arxiv.org/abs/0807.1960) as a good source for a more detailed discussion.
A simple case of a cluster algebra is the cluster algebra without frozen variables defined by a skew-symmetric matrix. A skew-symmetric matrix is essentially the same data (modulo some technicalities) as a quiver (directed graph) without loops or $2$-cycles, e.g.
$$\begin{pmatrix}0&-2\\2&0\end{pmatrix}\ \longleftrightarrow\ 1\Rightarrow 2$$
where the '$\Rightarrow$' represents a pair of arrows. The matrix here is the skew-symmetrisation of the adjacency matrix of the quiver, the assumptions on loops and $2$-cycles ensuring that there is no cancellation when computing this matrix, so that the process is invertible. (For simplicity, to avoid algebras breaking up into direct products, I consider only the pairs in the above correspondence such that the quiver is connected; this is most important in the classification statements below.) Such quivers, when they have no oriented cycles, also define associative algebras; they correspond to the finite dimensional basic hereditary algebras over a field $k$ via $Q\leftrightarrow kQ$ (I take $k$ to be algebraically closed, just in case). The algebra $kQ$ has basis given by the paths of $Q$ (including the length $0$ paths at each vertex) with multiplication given by concatenation of paths when the result is a path and $0$ otherwise, extended via linearity.
A classical result of Gabriel says that the algebra $kQ$ has finitely many isomorphism classes of indecomposable modules if and only if $Q$ is an orientation of one of the simply laced Dynkin graphs, i.e. those of type $\mathsf{A}_n$, $\mathsf{D}_n$ or $\mathsf{E}_{6,7,8}$. Suggestively, Fomin–Zelevinsky prove in their second paper on cluster algebras that the cluster algebras (without frozen variables, defined by a skew-symmetric matrix) with finitely many cluster variables are precisely those from matrices corresponding to these quivers.
It turns out that, whenever $Q$ is an acyclic quiver, there is in fact a bijection between most of the cluster variables (precisely, those not appearing in the initial seed) of the cluster algebra $\mathcal{A}_Q$ given by $Q$, and the indecomposable representations of $kQ$.
To make this connection stronger, one can introduce the 'cluster category' $\mathcal{C}_Q$, defined by Buan, Marsh, Reineke, Reiten and Todorov (https://arxiv.org/abs/math/0402054), which is a kind of extension of the category of $kQ$-modules by adding in some extra indecomposable objects corresponding to the initial cluster variables of $\mathcal{A}_Q$, so that each cluster variable $x$ of $\mathcal{A}_Q$ now corresponds to an indecomposable object $M_x$ of $\mathcal{C}_Q$.
Various properties of the cluster variables may now be translated into properties of these objects. For example, two cluster variables $x$ and $y$ are compatible (appear in the same cluster) if and only if the corresponding objects have the homological property $\operatorname{Ext}^1_{\mathcal{C}_Q}(M_x,M_y)=0$. The seeds of $\mathcal{A}_Q$ correspond to basic cluster-tilting objects of $\mathcal{C}_Q$, which are direct sums $T$ of some of the $M_x$s such that $\operatorname{Ext}^1_{\mathcal{C}_Q}(T,M_x)=0$ if and only if $M_x$ appears as a summand of $T$.
The quiver of the seed corresponding to such an object $T$ is the ordinary quiver of the endomorphism algebra $\operatorname{End}_{\mathcal{C}_Q}(T)$, i.e. the unique (up to isomorphism) quiver $Q_T$ such that $\operatorname{End}_{\mathcal{C}_Q}(T)\cong kQ_T/I$ for some ideal generated by linear combinations of paths of length at least $2$. The category $\mathcal{C}_Q$ is a '$2$-Calabi–Yau triangulated category', so its cluster-tilting objects have a mutation theory by work of Iyama–Yoshino (https://arxiv.org/abs/math/0607736) which turns out to correspond to mutations of seeds.
There is considerably more work in this direction, including cluster categories for cluster algebras corresponding to non-acyclic quivers by Amiot (https://arxiv.org/abs/0805.1035), and for cluster algebras with frozen variables by, e.g. Geiß–Leclerc–Schröer (https://arxiv.org/abs/math/0609138), Jensen–King–Su (https://arxiv.org/abs/1309.7301), Demonet–Iyama (https://arxiv.org/abs/1503.02362) and myself (https://arxiv.org/abs/1510.06224, https://arxiv.org/abs/1702.05352). One can also develop some theory for skew-symmetrizable matrices, see e.g. Demonet (https://arxiv.org/abs/0909.1633), Labardini-Fragoso–Zelevinsky (https://arxiv.org/abs/1306.3495) and Geiß–Leclerc–Schröer again (https://arxiv.org/abs/1410.1403).
While construction of these kinds of representation-theoretic models is useful for understanding cluster algebras, it also suggests lots of interesting representation theory that may not otherwise have been considered. This can make a precise connection difficult to pin down: I am not really aware of cluster algebras being used directly to solve open representation-theoretic problems, but they have suggested lots of new lines of representation-theoretic inquiry (e.g. the second reference to Geiß–Leclerc–Schröer above, which I think is not really about cluster algebras at all, a priori).
One example of this is Adachi–Iyama–Reiten's $\tau$-tilting theory (https://arxiv.org/abs/1210.1036). I am far from an expert on this theory, but my understanding is that this is really a purely representation-theoretic theory that could have been developed entirely without cluster algebras (and, I am told, was almost discovered by Auslander and co-authors in the 80s!) but in the end it was, at least in part, thinking about cluster categories that provided the inspiration.
I note that one slightly confusing part of this connection is that one quite rarely studies the cluster algebra itself as an algebra! There are some exceptions to this, such as work of Lampe (https://arxiv.org/abs/1210.1502).


TITLE: Electric Power and Energy
QUESTION [0 upvotes]: I was asked a question three times this week and I have been giving the same explanation over and over again, but I am starting to question myself.
if you have data on a PV panel production in the following form:
every 15min I have an instantaneous reading, so let's say that between 10am and 11:30am (1.5h) I had the following readings:

hour
reading

10:00
0.50 kW

10:15
0.75 kW

10:30
0.75 kW

10:45
0.50 kW

11:00
1.00 kW

11:15
0.75 kW

11:30
1.25 kW

What would be the total energy produced? and total power?
I am struggling to grasp the concept of measuring energy power at a specific time step, or measuring kWh every 30 min for example, if energy is power x time, how can I measure energy at a 30 min interval and how should I aggragate it for a day?
thanks all

REPLY [2 votes]: Total power produced is not a concept.  It would be like asking for "total speed" of a car.  You don't total rates.  You could have an average speed over a period of time, but not a total.
You can total energy, but only by making assumptions about what's happening during the times you didn't record the instantaneous power.  If the power changes, the total energy changes, but you might not notice.
Power multiplied by time is energy.  So if the readings you have are representative of that 15 minute period, you can approximate the total energy produced.  You'd multiply by that quarter hour.

Time
Power
15 min Energy
Total Energy

10:00
0.5 kW
0.125 kWh
0.125 kWh

10:15
0.75 kW
0.1875 kWh
0.3125 kWh

10:30
0.75 kW
0.1875 kWh
0.500 kWh

10:45
0.5 kW
0.125 kWh
0.625 kWh

11:00
1.0 kW
0.250 kWh
0.875 kWh

11:15
0.75 kW
0.1875 kWh
1.0625 kWh

11:30
1.25 kW
0.3125 kWh
1.375 kWh

Of course this is like asking how far your car has gone when you just are told the speed at a few specific times.  The car needs to not change speed much for you to be able to get close.  If it stops and starts a lot, your calculations won't mean anything.


TITLE: Continuous functions on locally compact Hausdorff spaces
QUESTION [2 upvotes]: Show every closed Ideal in $C_0(X)$ for a locally compact Hausdorff space X is of the form $K_F=\{f\in C_0(X)\colon f\lvert_F \equiv 0\}$ for some closed subspace $F\subseteq X$.
Here is what I have so far:
Let $I$ be a closed ideal.
I tried to use the Gelfand Transform.
Consider the evaluation map $ev_x: C_0(X) \rightarrow \mathbb{C}$ defined by $ev_x(f)= f(x)$ for all $x \in F$. Then we have $Ker (ev_x)= K_F$
From here, I'm pretty much lost what to do to get the result desired.
Also, I thought about using Urysohn's Lemma here: https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf . But still didn't anywhere.
Any help will be appreciated.

REPLY [3 votes]: Given a closed ideal $I\trianglelefteq  C_0(X)$, let
$$
  F=\{x\in X: f(x)=0,  \text{ for all } f\in I\},
  $$
and let us prove that $I=K_F$.
The inclusion  $I\subseteq K_F$ holds for a pretty elementary reason (which nevertheless sounds a bit like a tongue-twister):
every function in $I$ vanishes on any point where all functions in $I$ vanish.
The hard part is to prove the reverse inclusion $K_F\subseteq I$.   For this  pick  $f\in  K_f$.  Fixing  $\varepsilon >0$, recall that the set
$$
  C=\{x\in  X: |f(x)|\geq  \varepsilon \}
  $$
is compact.
Given  $x$ in $C$, we have that $f(x)$ is nonzero, so clearly $x$ is not in $F$.  Consequently not all functions in $I$
vanish at $x$, and hence we may choose  some element $a_x\in  I$, such that $a_x(x)\neq 0$.  The set
$$
  U_x=\{y\in X: a_x(y)\neq 0\}
  $$
is then an open neighborhood of $x$, and hence $\{U_x\}_{x\in C}$ is an open cover of $C$.  By compactness we may pick a
finite subcover $\{U_{x_1},U_{x_2},\ldots ,U_{x_n}\}$.  Next observe that
$$
  g:= \sum_{i=1}^n\overline {a}_{x_i}a_{x_i}
  $$
lies in $I$, and does not vanish on any point of $C$.
There are multiple endings for this story,  and perpahs the slickest way is as follows:  besides an ideal,  $I$ is a
C*-algebra and $g$ is a non-negative element in $I$, so $g^{1/p}$ lies in $I$  for every positive integer $p$.
Moreover the sequence $\{g^{1/p}\}_{p\in {\mathbb N}}$ converges uniformly to 1 on $C$.  It is also easy to see that
$$
  \|g^{1/p}\| =   \|g\|^{1/p} \leq  2,
  $$
for large enough $p$.
Therefore we may find  $p$ such that
$$
  |f(x)-g(x)^{1/p}f(x)| < \varepsilon , \quad\forall x\in  C,
  $$
and
$$
  |f(x)-g(x)^{1/p}f(x)|  \leq  \big (1+|g(x)^{1/p}|\big ) |f(x)|\leq  (1+2)\varepsilon ,
  $$
for every  $x\in  X\setminus C$.
Consequently $\|f - g^{1/p}f\|\leq  3\varepsilon $, so we deduce that $f$ lies in the closure of $I$, and hence also in $I$
because $I$ is closed.

EDIT. Here is another approach,  assuming we already know the answer in case $X$ is compact.
Assuming that $X$ is not compact, let $\hat X=X\cup \{\infty \}$
be the one-point compactification of $X$, and let
$$
  f\in C_0(X)\mapsto \hat f\in  C(\hat X)
  $$
be the natural identification.  One may then prove that $\hat I$ is a closed ideal in $C(\hat X)$, and hence, by the
compact case,
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\},
  $$
where $G$ is a closed subset of $\hat X$.
Notice that necessarily $\infty \in G$, since otherwise one could use Urysohn to
produce some $g\in C(\hat X)$, vanishing on $G$, with $g(\infty )\neq 0$, but this would imply that $g$ lies in
$\hat I\setminus C_0(X)$, which is impossible.
Setting $F=G\cap X$, we have  that $F$ is a closed subset of $X$.   So, upon identifying $C_0(X)$ as the subalgebra of $C(\hat
X)$ formed by the functions vanishing at $\infty $, we have:
$$
  \hat I=\{g\in C(\hat X) : g|_G=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  \{g\in C(\hat X) : g(\infty )=0\} = $$$$ =
  \{g\in C(\hat X) : g|_F=0\} \cap  C_0(X) = $$$$ =
  \{g\in C_0(X) : g|_F=0\} = K_F.
  $$


TITLE: Stuck on a problem in Rosen's book on number theory in function fields.
QUESTION [1 upvotes]: If $p$ is a prime number and $P \in \mathbb{F}_p[T] = A$ is irreducible, show $(A/ \langle P^2\rangle)^*$ is cyclic if and only if $\deg(P) = 1$.
I'm stuck on the $(\implies)$ direction.  Let $(A/ \langle P^2\rangle)^* = \langle f \rangle$, $n = \deg(P)$.  By the analogue of Fermat's little theorem and the fact that $f$ generates a group of order $p^n(p^n-1)$, we have that
\begin{align*}
f^{p^n - 1} \equiv 1 &\mod P
\\
f^{p^n} \equiv f &\mod P
\\
f^{p^n(p^n-1)} \equiv 1 &\mod P^2.
\end{align*}
I feel like this is what should give me the result since (unless I am mistaken) these three congruencies only happen when $(A/\langle P^d \rangle)^*$ is cyclic with $d=2$ (this is because $|(A/\langle P^d \rangle)^*| = p^{n(d-1)}(p^n - 1)$).  Is this the right idea?

REPLY [1 votes]: Let $\deg(P)=n$. Then $A/(P)\cong{\Bbb F}_{p^n}$ because $A/(P)$ is an extension of $\Bbb F_p$ of degree $n$.
Now if $(A/(P^2))^*$ is cyclic, let $f$ be a generator. We can write $f$ as $f=f_1+f_2P$ by Euclidean division, where $\deg(f_1)<n$. Then $$f_1^{p^n-1}=1+gP$$ since $\Bbb F_{p^n}^*$ is of order $p^n-1$.
Note that $f^{p^n-1}=f_1^{p^n-1}+hP=1+h'P$ for some $h,\,h'$. Thus
$$f^{p(p^n-1)}=(1+h'P)^p=1+(h'P)^p=1,$$
where the second equality is because $(a+b)^p=a^p+b^p$ in $\Bbb F_p[T]$, and the third because $p\geq2$.
This means the order of $f$ must divide $p(p^n-1)$. But the order of $A/(P^2)$ is $p^n(p^n-1)$ as you observed, so we conclude that $n\leq1$.

Hope this helps.


TITLE: Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$
QUESTION [4 upvotes]: How to evaluate this integral?
$$\int_{ - \infty }^\infty  {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$

Maybe we can start by $$\int_0^\infty  {\frac{{dx}}{{({x^2} + 1)\cosh ax}}}  = \frac{1}{2}\left[ {{\psi _0}\left( {\frac{a}{{2\pi }} + \frac{3}{4}} \right) - {\psi _0}\left( {\frac{a}{{2\pi }} + \frac{1}{4}} \right)} \right]$$ in this. Then take the derivative with respect to $a$, but I'm failed to solve it!

REPLY [3 votes]: $$I=\frac{1}{8} \int_{-\infty}^{\infty}
\frac{\cosh(7\pi z)}{(1+z^2)\cosh^3(3\pi z)} \text{d}z$$
Now we're considering the function
$$f(z)=\frac{\cosh(7\pi z)\psi^{(0)}\left ( 1-iz\right ) }{\cosh^3(3\pi z)}$$
From the residue theorem and calculate the residues at $$z=\frac{i}{6},\frac{i}{2},\frac{5i}{6}$$
We finally obtain
$$
\int_{-\infty}^{\infty} \frac{e^{7\pi z}}
{(e^{3\pi z}+e^{-3\pi z})^3(1+z^2)} \text{d}z
=\frac{3375\pi^3-12000\sqrt{3}\pi^2+25760\pi+26784\sqrt{3}  }{27000\pi^2}
$$


TITLE: higher Eilenberg-Moore-toposes of left exact derived comonads
QUESTION [4 upvotes]: It is a classical fact (e.g. here) that for an (accessible) right adjoint comonad on a (sheaf) topos, its Eilenberg-Moore category of coalgebras is itself a (sheaf) topos.
I suppose this remains true for $\infty$-toposes, for hypercomplete $\infty$-stack $\infty$-toposes at least? (But not assuming that the comonad is idempotent.)

REPLY [3 votes]: If $E$ is an ∞-topos and $T: E \to E$ is an accessible left exact comonad, then indeed the ∞-category $E^T$ of $T$-coalgebras is an ∞-topos. Moreover, it is hypercomplete if $E$ is. I will first show:

Lemma. Let $E'$ be presentable and let $U:E'\to E$ be a conservative functor that preserves colimits and pullbacks. Then $E'$ is an ∞-topos.

Proof. Recall that a presentable ∞-category $E'$ is an ∞-topos iff it satisfies descent, which means the following: If $a,b: K^\triangleright \to E'$ are cones, $\phi: a \Rightarrow b$ is a natural transformation, $b$ is a colimit, and $\phi$ is cartesian on $K$, then $a$ is a colimit iff $\phi$ is cartesian. In our situation, by the hypotheses on $U$, $a$ is a colimit iff $Ua$ is a colimit iff $U\phi$ is cartesian iff $\phi$ is cartesian. //
If $T$ is an accessible comonad on $E$, $E^T$ is accessible since it can be written as a limit of accessible ∞-categories. Moreover, the forgetful functor $U: E^T\to E$ creates colimits and is conservative ([1] 3.2.2.5 and 3.2.2.6). If $T$ is left exact, then $U$ is also left exact ([2] Corollary 5.5).
So the lemma implies that $E^T$ is an ∞-topos. Moreover, the forgetful functor $U$ is the left adjoint of a geometric morphism. Since $U$ preserves connectivity and is conservative, $E'$ is hypercomplete if $E$ is.
[1] J. Lurie, Higher Algebra
[2] E. Riehl and D. Verity, Completeness results for quasi-categories of algebras, homotopy limits, and related general constructions


TITLE: Hot air balloon trajectory predictions
QUESTION [0 upvotes]: Hey there Physics friends!
I've been struggling with what seems to be a very basic physics question. Let's say that I have wind vector for a given altitude and location, and I want to calculate the flight path of a hot air balloon.
So once the balloon hits that wind, would you calculate forward distance based on the wind speed (so would the balloon speed be exactly the average wind speed) or would you use the standard drag equation to calculate the force of the wind on the balloon to determine the distance traveled by the balloon in a given time period?

REPLY [1 votes]: Yes you would have to use the standard drag equation. Imagine the balloon is at rest (or moving with a different velocity from the wind) initially. As the balloon hits the wind it can't suddenly reach the same velocity as the balloon, as this would require an infinite acceleration (which is impossible). Instead the wind applies a force to the balloon. This force will increase the balloon's velocity until it does match that of the wind (drag force is proportional to the relative speed between the balloons), at which point the velocity of the balloon will match that of the wind. If you can assume that the time it takes for the balloon to reach the velocity of the wind is negligible (such as in the case of a large drag coefficient or cross sectional area of the balloon) then the velocity of the balloon could be assumed to be the same as that of the wind at all times.


TITLE: Gelfand-Mazur complex part mean what it says?
QUESTION [0 upvotes]: This question pertains to the complex Gelfand/Mazur theorem.
Mazur/Gelfand says normed division algebra over the complex numbers is isomorphic to the complex numbers.
Therefore, is the statement true: The ONLY normed division algebra over complex numbers is the complex numbers? And, therefore, is this statement true: a normed division algebra over complex numbers is not isomorphic to the reals or quaternions. (There is another post here that appears to confirm the quaternion side of this question but not for the real side.)
Some argue that this was not what the theorem meant because the complex numbers can be seen to contain the reals. But that would seem to require the theorem to say that a normed division algebra over the complex numbers is isomorphic to the reals and complex or some kind of transitivity group argument would then put it isomorphic with quaternions (which does not seem to be right.)
Clarification: this question is NOT about normed division algebra over real numbers but about normed division algebra over complex.

REPLY [1 votes]: For Banach Algebra, Wikipedia states, "The only complex Banach algebra which is a division algebra is the complexes. (This is known as the Gelfand–Mazur theorem.)" The "only" was the qualifier this question was asking to be confirmed. (Apologies for asking such a fundamental question.) Also, "Unital Banach algebras over the complex field provide a general setting to develop spectral theory."
Rickart, C. E., An elementary proof of a fundamental theorem in the theory of Banach algebras,  says, "Mazur-Gelfand ... is directly equivalent to the existence of a spectrum for elements of a normed algebra."
Therefore, $\mathbb{R}$ and $\mathbb{H}$ are not complex Banach algebras. If given a Banach algebra over the complex then it is only isomorphic to $\mathbb{C}$.


TITLE: Is this festive picture a valid equation?
QUESTION [2 upvotes]: Ive seen this picture floating around, and I was wondering if it was a legitimate mathematical equation, or if it just looks convincing enough to fool Luddites like myself?

REPLY [3 votes]: They're correct (with a few caveats that others have pointed out).
The manipulations:

Original equation
Multiply both sides by $r^2$
Exponentiate both sides (make each side the exponent base $e$).  The exponent doesn't show up on the right side because $e^{\log_e(f(x))} = f(x)$.
Multiply both sides by $m$.
Write out $r^2$ as $rr$ (both mean $r \cdot r$) and $msa$ as $mas$ and $yrr$ as $rry$ (valid by commutative property of multiplication).


TITLE: Taylor development on manifolds and Manifolds of differentiable Mappings?
QUESTION [1 upvotes]: I'm trying to study the book "Manifolds of Differentiable Mappings" written by Michor and I came across the following:
He considers two smooth manifolds $M$ and $N$ and define an equivalence relation $\simeq$ on $C^\infty(M,N)\times M$ as follows:
$(f, p)\simeq (g, q)$ if and only if $p=q$ and $f$ and $g$ have the same Taylor development of order $k$ at $p$ in some pair of coordinate charts about $p$ and $f(p)$, respectively.
Can anyone explain me this?
More precisely, how is the Taylor development of order $k$ at $p$ is define for a function $f:M\longrightarrow N$ between manifolds?
References will be welcome..
Thanks

REPLY [2 votes]: You don't need to define the Taylor development for functions $M \to N$ since you're working in the coordinate charts he mentions in the definition. For example if the coordinate chart about $p$ is $\phi : U \subset M \to \mathbb R^m$ and likewise $\psi : V \subset N \to \mathbb R^n$ about $f(p)$ then he is talking about the Taylor polynomials of the maps $$\psi \circ f  \circ\phi^{-1}, \psi \circ g \circ \phi^{-1} : \phi(U) \to \mathbb R^n.$$ These are just functions between Euclidean spaces, so you can take the component-wise Taylor expansion as usual.


TITLE: Dense subset relation
QUESTION [0 upvotes]: Defn Let $B$ be a Boolean algebra. A subset $D$ of $B$ is called b-dense if for every $0\neq b\in B$, there is $0\neq d\in D$ such that $d\leq b$.
Defn Let $T$ be a topological space. A subset $D$ of $T$ is called t-dense if the closure $Cl(D)=T$.
Let $T=Stone(B)$, the Stone space of $B$. I am asking what is the relation between b-dense in $B$ and t-dense in $T$.
Any idea ??

REPLY [0 votes]: Dense subsets of a Boolean algebra correspond to $π$-bases consisting of clopen subsets of corresponding Stone space.

A collection of non-empty open subsets of a topological space is called $π$-base if every non-empty open set of the space contains some member of the collection as a subset.

$π$-bases are somewhere between bases and dense subsets. Each base is a $π$-base and union of $π$-base (or choosing a point in each set) is a dense set.


TITLE: Expressing vectors in different bases
QUESTION [1 upvotes]: I am trying to wrap my head around the concept of expressing vectors in different bases, and how to 'translate' between them.
Consider the two bases given by:
$$B={e_1,e_2,e_3},\quad B^{\prime}=\left\{\left[\begin{array}{l}
1 \\0 \\0\end{array}\right],\left[\begin{array}{l}1 \\1 \\0
\end{array}\right],\left[\begin{array}{l}
1 \\1 \\1\end{array}\right]\right\}$$
If I understood correctly, the columns of $B'$ also determine the change of basis matrix:
$$P=\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right]$$
So that the basis vectors in $B$ can now be expressed in terms of the basis vectors in $B'$ (sorry for a bit sloppy notation):
$$b_{j}=\sum_{i} p_{i j} b_{i}^{\prime}, \qquad \{b_{j}\} = B, \quad \{b_{i}^{\prime}\} = B^{\prime}$$
Which in matrix notation becomes:
$$b=Pb'$$
For example, if we have a vector,$v'$, that in basis $B'$ has coordinates $(1,4,2)$, then in $B$ the same vector will have coordinates:
$$v=P v^{\prime}=\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
4
\end{array}\right]=\left[\begin{array}{l}
7 \\
6 \\
4
\end{array}\right]$$
Is there anything that I misunderstood or does this look correct?

REPLY [1 votes]: The matrix $P$ is the matrix that changes $B'$-coordinates into $B$-coordinates. Note that $P$ arises as the matrix representantion of the identity transformation $Id:\mathbb{R}^3 \to \mathbb{R}^3$ with respect to the bases $B'$ and $B$, respectively. This is really easy to see, since the elements of $B'$ are already written in $B$-coordinates.
So, to obtain the expression in $B$-coordinates of any vector $v = a_1b'_1 + a_2b'_2 + a_3b'_3$ written in $B'$-coordinates you use multiplication by $P$
$$[v]_B = P \begin{bmatrix}
a_1\\
a_2\\
a_3
\end{bmatrix}$$
The coordinates of $[v]_B$ give you the coefficients of $v$ in $B$-coordinates. Conversely, if you wanted to obtain $B'$-coordinates from $B$-coordinates you would have to carry out the same procedure using $P^{-1}$ instead.


TITLE: Jacobian with vanishing determinant everywhere
QUESTION [7 upvotes]: This is from advanced calculus.
Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable. The Jacobian of $f$ has zero determinant over $\mathbb{R}^n$. For example in 2-dimensional case, $f(x,y)=(y,y)$ is such a function. Let's suppose $n \ge 2$ since $n=1$ case gives a constant function.
My question is :is it true that $f$ cannot be 1-1?
My best attempt is that at each $x \in \mathbb{R}^n$ some directional derivative is zero, and perhaps an integral over some path will work, but I am not sure how to do it.
Any idea is welcome. Thank you very much.

REPLY [3 votes]: There are (at least) two ways to prove this:
The first is "easy", but uses theorems which are more difficult: Assume that $f$ is 1-1. Then, by invariance of domain, it follows that $f$ is an open map, i.e., if $U \subset \Bbb{R}^n$ is open, then so is $f(U)$. In particular, $f(\Bbb{R}^n)$ is nonempty and open and thus has positive Lebesgue measure. But Sard's theorem shows that $f(\Bbb{R}^n)$ is a set of measure zero, contradiction.

For a more elementary proof, one might argue as follows: Let $k := \max_{x \in \Bbb{R}^n} \text{rank}(Df(x))$ be the maximal rank of $f$. It is then not too hard to see that $U := \{x \,:\, \text{rank}(Df(x)) = k\}$ is open and nonempty.
Now, the constant rank theorem (see e.g. here) shows for arbitrary $x_0 \in U$ that there is a neighorhood $V$ of $x$ and a neighborhood $W$ of $f(x)$ such that $f(V) \subset W$ and there are $C^1$ diffeomorphisms $\Phi : W \to W'$ and $\Psi : V \to V'$ such that
$$
(\Phi \circ f \circ \Psi^{-1})(x_1, \dots, x_n) = (x_1, \dots, x_k, 0,...0)
$$
for all $(x_1, \dots, x_n) \in V'$. Hence, $\Phi \circ f \circ \Psi^{-1}$ is not 1-1, which easily shows that $f$ is not 1-1 either.


TITLE: Prove concavity of a particular function
QUESTION [2 upvotes]: The exercise is the following. Let $f:U\longrightarrow\mathbb{R}$ be semiconcave on the open set $U$, that is there exists a constant $K\geq0$ such that
$$
\lambda f(x)+(1-\lambda)f(y)\leq f(\lambda x+(1-\lambda)y)+\frac{1}{2}K\lambda(1-\lambda)|x-y|^2
$$
for all $x, y\in U$ and $\lambda\in[0, 1]$. Prove that the above inequality is equivalent to the concavity of the map $x\mapsto f(x)-\frac{1}{2}K|x|^2$. I cannot prove this equivalence. In particular, by the concavity of $x\mapsto f(x)-\frac{1}{2}K|x|^2$, I obtain the inequality
$$
\lambda f(x)+(1-\lambda)f(y)\leq f(\lambda x+(1-\lambda)y)+\frac{K}{2}(\lambda|x|^2+(1-\lambda)|y|^2-|\lambda x+(1-\lambda)y|^2).
$$
How can I get the desired result?
Thank You

REPLY [1 votes]: Expanding
$$\lvert \lambda x + (1 - \lambda y)\rvert^2 = \lambda^2 \lvert x\rvert^2 + 2\lambda(1 - \lambda)\langle x,y\rangle + (1 - \lambda)^2 \lvert y\rvert^2$$
you can write
$$\lambda\lvert x\rvert^2 + (1 - \lambda)\lvert y\rvert^2 - \lvert \lambda x + (1 - \lambda) y\rvert^2 = (\lambda - \lambda^2)\lvert x\rvert^2 - 2\lambda(1 - \lambda)\langle x,y\rangle + [(1 -\lambda) - (1 - \lambda)^2]\lvert y\rvert^2$$ or $$\lambda(1 - \lambda)\lvert x\rvert^2 - 2\lambda(1 - \lambda)\langle x,y\rangle + (1 - \lambda)\lambda \lvert y\rvert^2$$ The latter expression is the same as $$\lambda(1 - \lambda)\lvert x - y\rvert^2$$
You can do something similar for the proof of the other direction.


TITLE: Poisson processes understanding solution waiting time
QUESTION [2 upvotes]: I am trying to understand the solution for the following example:
Assume that passengers arrive at a bust station as a Poisson process with rate $\lambda$. a) The only bus departs after a deterministic time $T$. Let $W$ be the combined waiting time for all passengers, compute $E(W)$.
The solution says the following: If $S_1, S_2, ... $ are the arrival times in $[0,T]$, then the combined waiting time is $W = T- S_1 + T-S_2 +...$. Let $N(T)$ be the number of arrivals in $[0,t]$. We obtain the answer by conditioning on the value of $N(T)$. WE have $$E(W) = \sum_{k=0}^{\infty} E(W|N(T)=k)P(N(T)=k) = \sum_{k=0}^{\infty} k\frac{T}{2} P(N(T)=k) = \frac{T}{2}E(N(T)) $$ My question comes in the second equality, why is $E(W|N(T)=k)= k\frac{T}{2}$?  I understan the expression would be $E(W|N(T)=k) = E(\sum_{l=1}^{N(T)}T-S_i|N(T)=k) = kE(T-S_i)$ but then I don't seehow the last expression becomes just $\frac{T}{2}$. Thanks for your help!

REPLY [1 votes]: $T$ is deteministic so you have
$$E[T-S_i] =T-E[S_i]$$
It works if you use the assumption that $S_i$ are uniformally distributed in $[0,T]$ in order to get $E[S_i]=T/2$.
Edit
It is indeed stated in your paper, under thm 18.5 Uniformity of previous event time :
Given that $N(t) = k$, the conditional distribution of the interarrival times, $S_1,\ldots,Sk$, is distributed as order statistics of $k$ independent uniform variables. The set $$\{S_1,\ldots, S_k\} =\{U_1,\ldots, U_i\}$$, where $U_i$ are independent and uniform on $[0, t]$.


TITLE: "Different reality" inside a black hole
QUESTION [0 upvotes]: why does our perception of space and time entirely change inside a black hole? And why does time not stop inside it from the perspective of the inside observer, however, extremely slows down for the outside observer?

REPLY [1 votes]: Our perception of space and time would not change inside a black hole (if by that you mean inside the event horizon).
A person falling into a black hole would cross the event horizon smoothly and would not notice anything untoward at all unless the black hole was of low enough mass that the effective tidal forces across their body become large enough to notice. In particular - proper time, the time recorded on their own watch, continues to proceed in a forward direction at exactly the same experiential rate.
Perhaps what they would notice is that it would prove impossible for them to move towards a larger radial coordinate no matter what force was applied. They would also start to see strange optical effects/distortions associated with the extreme curvature of spacetime at small radii. There would then only be a finite amount of (proper) time before they were annihilated at the singularity, but the time interval between feeling uncomfortable and being annihilated would be less than 1 second (whatever the mass of the black hole).
I think what you are referring to is the oft-mentioned switch in the roles of the timelike/spacelike character of the $r$ and $t$ coordinates in the Schwarzschild coordinate system. All this means is that in order to ensure that proper time continues to increase then your $r$ coordinate must decrease when beneath the event horizon. In other words, the $r$ coordinate can only change in one direction and thus moving towards smaller $r$ becomes as inevitable as the passage of time in a forward direction in the exterior universe.


TITLE: Simple question about full time derivative
QUESTION [0 upvotes]: Let's have full time derivative equation
$$
\frac{F(\mathbf r (t), \mathbf p (t), t)}{dt} = \frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} + \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } + \frac{\partial F}{\partial t}.
$$
Can I represent this equation as
$$
F(\mathbf r (t), \mathbf p (t), t) = \int \left( \frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} + \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } + \frac{\partial F}{\partial t}\right)dt
$$
and, if I can't, why not?
In principle, this expression is equivalent to the identity
$$
F = \int \frac{dF}{dt}dt.
$$

REPLY [1 votes]: You can write
$$dF= \frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} dt+ \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } dt+ \frac{\partial F}{\partial t}dt$$
$$dF= \bigg(\frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} + \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } + \frac{\partial F}{\partial t}\bigg)dt$$
From which it follows that
$$\frac{dF}{dt}= \frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} + \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } + \frac{\partial F}{\partial t}$$
And since
$$F=\int dF=\int \bigg(\frac{\partial \mathbf r }{\partial t}\frac{\partial F}{\partial \mathbf r} + \frac{\partial \mathbf p }{\partial t}\frac{\partial F}{\partial \mathbf p } + \frac{\partial F}{\partial t}\bigg)dt=\int\frac{dF}{dt}dt$$


TITLE: Maximal ergodic inequality
QUESTION [4 upvotes]: A map $f: X \to X$ preserves an ergodic probability $\mu$, i.e., $\mu \circ f^{-1}=\mu$ and for any $\phi: X \to \mathbb{R}$ with $\int \phi d\mu=0$,
$$\frac{1}{n} \sum_{i \le n} \phi \circ f^i \to 0 \text{ almost surely and in } L^1(\mu).$$
Therefore, $\max_{n \ge N} \frac{1}{n} \sum_{i \le n} \phi \circ f^i \to 0$ almost surely.
Similar to maximal inequality, are there references to quantitatively study
$$\max_{n \ge N} \frac{1}{n} \sum_{i \le n} \phi \circ f^i \text{ ?}$$
Similar to martingale inequality, are there references to study
$$\lVert\max_{n \ge N} \frac{1}{n} \sum_{i \le n} \phi \circ f^i\rVert_{L^1} \precsim  \lVert\frac{1}{N} \sum_{i \le N} \phi \circ f^i\rVert_{L^1} \text{ ?}$$

REPLY [4 votes]: For a nice introductory discussion to the maximal ergodic inequality, see
[1].  In particular,   inequality (5) there is Wiener's maximal ergodic theorem. See also Lemma 15.3 in [2]. A more advanced account, that in particular includes the $L^p$ maximal inequalities, is in the book [3], see Cor 2.2 page 8 for the Wiener inequality and the proof of Theorem 6.3 page 52 for the $L^p$ inequality
$$\lVert\max_{n \le N} \frac{1}{n} \sum_{i \le n} \phi \circ f^i\rVert_{L^p} \le  \frac{p}{p-1} \lVert \phi \rVert_{L^p} \quad (*) $$
valid for $p>1$. Note that Theorem 6.3 in [3] only states the limiting inequality when $n$ tends to infinity on both sides of $(*)$, but the proof gives $(*)$ for every finite $n$ as well, since it is a direct application of Cor. 2.2 and Lemma 6.2.
For $p=1$ the inequality (*) does not hold, one needs an extra log term on the RHS, see (6.6)on page 52 in [3].
Finally, note that a general inequality of the form proposed,
$$\lVert\max_{n \ge N} \frac{1}{n} \sum_{i \le n} \phi \circ f^i\rVert_{L^p} \precsim  \lVert\frac{1}{N} \sum_{i \le N} \phi \circ f^i\rVert_{L^p} \quad (**) $$
cannot be valid for any $p$. Consider, for example, $f$ to be a rotation of angle $2\pi(\frac{1}{N}-\epsilon)$ on the unit circle, where $\frac{1}{N}-\epsilon $ is irrational. Let $\phi$ be a function that equals 1 on an arc of length  $\pi$ on that circle, and equals $-1$ on the complementary arc of the same length. Then for fixed $N>1$, the RHS of $(**)$ tends to zero as $\epsilon \to 0$, while the LHS is at least $1/(N+1)$.
[1] http://www-stat.wharton.upenn.edu/~steele/Courses/530/Resources/HopfSLLN.pdf
[2] https://people.math.wisc.edu/~roch/teaching_files/275b.1.12w/lect15-web.pdf
[3] Krengel, Ulrich. Ergodic theorems. Vol. 6. Walter de Gruyter, 2011.
https://books.google.ca/books?hl=en&lr=&id=t4_BDh8gt2kC&oi=fnd&pg=PA1&ots=5R-dupukqq&sig=yx6aMMlq93oOvD4eLotHFdk6q7U&redir_esc=y#v=onepage&q&f=false


TITLE: How to solve differential equation for two-dimensional diffusion?
QUESTION [2 upvotes]: For solving the simplified one-dimensional Fick's equation
$$ \frac{\partial c}{\partial t} = D \frac{\partial^2 c}{\partial x^2} $$
with the common boundary conditions, the Laplace transform results in
$$c = \frac{A}{t^{1/2}}exp(-x^2/4Dt)$$
but if having a two-dimensional diffusion (assume diffusion over a flat surface) as
$$ \frac{\partial c}{\partial t} = D \Bigl(\frac{\partial^2 c}{\partial x^2} + \frac{\partial^2 c}{\partial y^2}\Bigl)$$
then, how will be the Laplace transform to solve the two-dimensional form of the Fick's second law?

REPLY [1 votes]: You can look up solutions to the heat equation in higher dimensions.
The equation $$ \partial_t u(\vec{x},t) = \kappa\sum_i \partial_{ii} u(\vec{x},t)  $$ has Green's function (fundamental solution)
$$ \Psi(\vec{x},t) = \frac{1}{\sqrt{(4\pi \kappa t)^n}}\exp\left(\frac{-[\vec{x}\cdot\vec{x}]}{4\kappa t}\right) $$
So that given initial conditions $u(\vec{x},0)=\mathcal{I}(\vec{x})$, the solution is $$ u(\vec{x},t)=\int_{\mathbb{R}^n} \Psi(\vec{x}-\vec{y},t)\;\mathcal{I}(\vec{y})\, d\vec{y} $$
Using the notation of the heat kernel instead, we can write: $$ \mathfrak{K}_t(\vec{x},\vec{y})=\frac{1}{\sqrt{(4\pi \kappa t)^n}}\exp\left(\frac{-||\vec{x}-\vec{y} ||^2_2}{4\kappa t}\right)  $$
so it has solution: $$ u(\vec{x},t)=\int_{\mathbb{R}^n} \mathfrak{K}_t(\vec{x},\vec{y})\;\mathcal{I}(\vec{y})\, d\vec{y} $$
So if $n=2$ and $\mathcal{I}(\vec{y})=\delta(\vec{y})$, we get:
$$
u(x,y,t) = \frac{1}{{4\pi \kappa t}}\exp\left(\frac{-[x^2+y^2]}{4\kappa t}\right)
$$


TITLE: Concordance classes of diffeomorphisms of $D^4$
QUESTION [4 upvotes]: Is anything known about the set of concordance classes (also called pseudoisotopy classes) of the relative to the boundary diffeomorphisms of $D^4$?

REPLY [5 votes]: It is the trivial group.
Firstly, by isotopy extension the extension-by-the-identity homomorphism to the group of pseudoisotopy classes of orientation-preserving diffeomorphisms of $S^4$,
$$\widetilde{\pi}_0(\mathrm{Diff}_\partial(D^4)) \to \widetilde{\pi}_0(\mathrm{Diff}^+(S^4)),$$
is surjective with kernel generated by the Dehn twist around the boundary. But the Dehn twist is isotopic to the identity for $D^4$ (it is easy to produce an explicit isotopy), so this homomorphism is an isomorphism.
Then for $S^4$ you just invoke Theorem 1 of Kreck's Isotopy classes of diffeomorphisms of $(k-1)$-connected almost-parallelizable $2k$-manifolds, which says that for any closed 1-connected smooth $4$-manifold $M$, the homomorphism
$$\widetilde{\pi}_0(\mathrm{Diff}(M)) \to \mathrm{Aut}(H_2(M;\mathbb{Z})),$$
induced by the action on homology, is injective.


text:
\section{The global exceptional isomorphism}
\label{sec:global-ex-isom}

In this section, we construct the global exceptional
isomorphism between a (projectivized) unitary group attached to
a hermitian (or skew-hermitian) space $\VV$ and the identity component of
a (projectivized) quaternionic unitary group attached to a
quaternionic {\it skew-hermitian} space $\tilde{V}$. Moreover,
we study the restriction of this isomorphism to certain natural
subgroups corresponding to the decomposition of $\VV$ into the direct sum of
two subspaces.

\subsection{Hermitian spaces and unitary groups}

Let $E/F$ be a quadratic extension and $\VV$ an $E$-hermitian space.
Thus $\VV$ is equipped with a non-degenerate form
\[
(\cdot,\cdot): \VV \times \VV \rightarrow E
\]
satisfying
\[
(v\alpha ,w\beta ) = \alpha^\rho (v,w) \beta, \quad (v,w)=(w,v)^\rho.
\]

\begin{prop}
\label{prop:norm-sim}
Let $\VV$ be a hermitian space over $E$ of dimension $n$ and let $g\in \GU_E(\VV)$. Then
\[
\N (\det(g)) = \nu(g)^n,
\]
where $\N$ denotes the norm map $E^\times \rightarrow F^\times$.
\end{prop}

\begin{proof}
This is obviously well known but the proof will serve to establish some notation.
Let  $\VV^*$ be the $E$-dual of $\VV$.
First, the form $(\cdot,\cdot)$ induces an $E$-conjugate linear isomorphism
\[
\varphi: \VV \simeq \VV^*,
\]
given by
\[
\varphi(x) (y) = ( x,y ).
\]
Then for $r$ a positive integer,
\begin{equation}
\label{eqn:wedge-d-phi}
\wedge^r \varphi: \wedge^r \VV \rightarrow \wedge^r \VV^*
\end{equation}
is also $E$-conjugate linear. Let $\iota$ be the $E$-linear isomorphism
\begin{equation}
\label{eqn:iota-defn}
\iota:  \wedge^r (\VV^*) \simeq (\wedge^r \VV)^*,
\end{equation}
 induced by the multilinear map
\begin{equation}
\label{eqn:multilinear-pairing}
(\VV^*)^r \times \VV^r \rightarrow E, \quad (\lambda_1, \ldots, \lambda_r, \v_1, \ldots, \v_r) \mapsto \det ( \lambda_i (\v_j)).
\end{equation}
Now any $g\in \GL_E(\VV)$ acts on $\VV^*$ via $g\lambda (\v) = \lambda (g^{-1}\v)$ and $\iota$ is equivariant for this action since
\eqref{eqn:multilinear-pairing} is equivariant for the diagonal action of $\GL_E (V)$. The composite
\[
\iota \circ \wedge^r \varphi: \wedge^r \VV \rightarrow (\wedge^r \VV)^*
\]
is an $E$-conjugate linear isomorphism and may be viewed as giving a hermitian form on $\wedge^r \VV$,
denoted $(\cdot, \cdot)_{\wedge^r \VV}$. (That this form is conjugate symmetric follows for instance by computing it in matrix form in terms of the matrix
of the form on $\VV$ with respect to an orthogonal basis. If the matrix of the original form is the diagonal matrix with entries $a_1, \ldots, a_n$, then  the entries $a_i$ lie in $F$ and the form on
$\wedge^r \VV$ is represented by the diagonal matrix whose entries are products of the form $a_{i_1} \cdots a_{i_r}$ with $1\le i_1 < \cdots < i_r \le n$. )

Now suppose $g\in \GU_E (\VV)$. Then
\[
\varphi (g\v) (\w) = (g\v,\w) =\nu(g)^{-1} (\v, g^{-1} \w) = \nu(g)^{-1} (g \varphi) (\w)
\]
so that
\[
\varphi \circ g = \nu(g)^{-1} g \circ \varphi
\]
and
\[
\wedge^r \varphi \circ g = \nu(g)^{-r} \cdot g \circ \wedge^r \varphi,
\]
so that for $\x,\y \in \wedge^r \VV$, we have
\[
(g \x, g \y)_{\wedge^r \VV} = \nu(g)^r (\x,\y)_{\wedge^r \VV}.
\]
Now take $r=n$. Then $g$ acts on $\wedge^n \VV$ as the scalar $\det(g)$, so that $(g\x, g\y)_{\wedge^n \VV} = \N (\det(g)) (\x,y)_{\wedge^n \VV}$, from
which it follows that $\N (\det(g)) = \nu(g)^n$.
\end{proof}

\subsection{Construction of the (global) exceptional isomorphism}
\label{sec:constr-gl-ex-isom}

Let $\VV$ be a four dimensional (right) $E$-hermitian space. Such a $\VV$ is classified at archimedean places by its signature and
at non-archimedean places by its determinant $\delta_v \in F^\times/\N E^\times$, which equals the discriminant (since $\dim \VV =4$.)
(For an archimedean place, $\delta_v=1$ if the signature is either $(4,0)$, $(2,2)$ or $(0,4)$ and
is nontrivial if the signature is $(3,1)$ or $(1,3)$. )
Let $B$ be the unique quaternion algebra over $F$ which is ramified exactly at those places
$v$ of $F$ at which $\delta_v$ is nontrivial. We will construct
\begin{itemize}
\item A three dimensional right $B$-space $\tilde{V}$.
\item A skew-hermitian $B$-form $\langle \cdot, \cdot \rangle$ on $\tilde{V}$.
\end{itemize}
such that there is a natural isogeny
\[
\GSU_E (\VV) \rightarrow \GU_B(\tilde{V})^0,
\]
where
\[
\GSU_E (\VV) = \{ g\in \GU_E(\VV): \ \det (g)=\nu(g)^2 \}
\]
with $\nu$ being the similitude norm, as well as a natural isomorphism
\[
\PGU_E( \VV) \simeq \PGU_B (\tilde{V})^\0.
\]

Let $\tilde{\VV} = \wedge^2 \VV$.
This is a right $E$-space and we will extend the $E$-action to
a right $B$-action. To do so, we must construct an element
$L\in \End_F (\tilde{\VV})$ which is conjugate linear for the $E$-action:
\[
L(x \alpha) = (Lx) \alpha^\rho
\]
for $x\in \tilde{\VV}$, $\alpha \in E$.

The map $L$ will be a composite of three maps:
\begin{enumerate}
\item   The map
\[
\wedge^2 \varphi: \wedge^2 \VV \rightarrow \wedge^2 (\VV^*)
\]
obtained by specializing \eqref{eqn:wedge-d-phi} to $d=2$, which is an $E$-conjugate linear isomorphism.

\item The map
\[
\iota: \wedge^2 (\VV^*) \simeq (\wedge^2 \VV)^*.
\]
obtained by specializing \eqref{eqn:iota-defn} to $d=2$, which is an $E$-linear isomorphism.

\item Here we use that $\dim \VV =4$.
Fix an isomorphism
\[
d: \wedge^4 \VV \simeq E.
\]
This is well defined up to scaling.
The natural map
\begin{equation}
\label{eqn:pairing-on-wedge-2}
\wedge^2 \VV \times \wedge^2 \VV \rightarrow \wedge^4 \VV \simeq E
\end{equation}
is {\it symmetric} and
induces an $E$-linear isomorphism
\[
\psi: \wedge^2 \VV \simeq (\wedge^2 \VV)^*.
\]

\end{enumerate}

Let
\[
L= \psi^{-1} \circ \iota \circ \wedge^2 \varphi.
\]
Clearly $L$ depends on the choice of $d$. If $d$ is scaled by $\alpha$, then $\psi$ is scaled by $\alpha$ as well and $L$ is scaled by $\alpha^{-1} $.
However $L^2$ changes to
\[
(\alpha^{-1} L)(\alpha^{-1} L) =  (\alpha^\rho \alpha)^{-1} L^2 = \N (\alpha)^{-1} L^2.
\]
Thus $L^2$ is well defined up to norms from $E^\times$ to $F^\times$.
In fact, $L^2$ turns out to be a scalar operator. To identify this scalar, we recall
the following invariant attached to a hermitian space $\VV$ of dimension $n$ and an
isomorphism $d: \wedge^n V \simeq E$.

\begin{defn}
Let $\VV$ be a hermitian space of dimension $n$ with form $H$ and let
$d: \wedge^n \VV \simeq E$ be an isomorphism. The form $H$ induces
a map
\[
\VV^n \times \VV^n \rightarrow E, \quad (\v_1, \ldots, \v_n, \w_1, \ldots, \w_n) \mapsto \det [ H(\v_i, \w_j ) ],
\]
which factors through $\wedge^n \VV \times \wedge^n \VV$
and gives a hermitian form
\[
h: \wedge^n \VV \times \wedge^n \VV \rightarrow E.
\]
Let $v \in \wedge^n \VV$ be such that $d(v)=1$. Then define
\[
\vol (H,d) = h (v,v).
\]
Note that $h$ is a hermitian form, so $\vol(H,d)$ lies in $F^\times$ and its class in $F^\times/\N E^\times$
equals the class of the determinant of $H$.
\end{defn}

\begin{prop}
\label{prop:volume-V-d}
The map $L^2$ is multiplication by $\vol ((\cdot,\cdot)_\VV,d)$.
\end{prop}

\begin{proof}
We will pick a suitable basis and compute.
Since $L^2$ and $\vol ((\cdot,\cdot)_\VV,d)$ scale in exactly the same way
as a function of $d$, we can choose any convenient $d$ as well.
Let $\v_1, \ldots, \v_4$ be
a basis of $\VV$ with respect to which the form $(\cdot, \cdot)_\VV$ is diagonal with entries
$a_1, \ldots, a_4 \in F$. Let $\e_1,\ldots, \e_4$ be the dual basis of $\VV^*$.
Then
\[
\varphi (\v_i ) = a_i \e_i
\]
and
\[
\wedge^2 (\varphi) (\v_i \wedge \v_j) = a_i a_j \e_i \wedge \e_j.
\]
For $1\le i < j \le 4$, let $\v_{ij}$ denote the element $\v_i \wedge \v_j \in \wedge^2 \VV$.
This collection gives a basis of $\wedge^2 \VV$. We let
$\{ \e_{ij} \} \subset (\wedge^2 \VV)^*$ be the dual basis.
Then
\[
\iota (\e_i \wedge \e_j) = \e_{ij}.
\]
For any pair $(i,j)$ as above let $(i',j')$ be the unique pair of elements such that $\{i,j,i',j'\} =\{ 1,2,3,4\}$ and such that
$i' < j'$. Define $\sign(i,j) =\pm 1$ by
\[
\v_{ij} \wedge \v_{i'j'}=\sign(i,j) \v_1 \wedge \v_2 \wedge \v_3 \wedge \v_4.
\]
Now {\it choose} $d$ such that
\begin{equation}
\label{eqn:choice-of-d}
d (\v_1 \wedge \v_2 \wedge \v_3 \wedge \v_4) = -1.
\end{equation}
(This choice may seem surprising but it is made so as to agree with some conventions in periods1.)
Then
\[
\psi^{-1} (\e_{ij}) =- \sign (i,j) \cdot \v_{i'j'}.
\]
(The $-$ sign here occurs because of the choice made in \eqref{eqn:choice-of-d}.)
Now we can write down $L$ explicitly in the basis $\v_{ij}$. It is given by:
\begin{align*}
\v_{12} & \mapsto -a_1 a_2 \v_{34} \\
\v_{13} & \mapsto a_1 a_3 \v_{24} \\
\v_{14} & \mapsto -a_1 a_4 \v_{23} \\
\v_{23} & \mapsto -a_2 a_3 \v_{14} \\
\v_{24} & \mapsto a_2 a_4 \v_{13} \\
\v_{34} & \mapsto -a_3 a_4 \v_{12}
\end{align*}
The proposition follows from this explicit description.
\end{proof}

Now let us define a quaternion algebra $B$ as follows. Let
\[
J:=\vol( (\cdot,\cdot)_\VV,d)
\]
and define $B$ by
\[
B := E + E\j, \quad \j^2 = J, \quad \alpha \j = \j \alpha^\rho
\]
for all $\alpha \in E$. Then we can define a right action of $B$ on $\tilde{\VV}$ by
\[
x \cdot \j = L(x).
\]
We will denote this space by $\tilde{V}$ when we want to regard it as a $B$-space rather than an $E$-space.

As in the proof of Prop. \ref{prop:norm-sim}, the composite map $\iota \circ \wedge^2 \varphi$ is a conjugate linear isomorphism
\[
\wedge^2 \VV \simeq (\wedge^2 \VV)^*
\]
that gives rise to a hermitian form $(\cdot, \cdot)_{\tilde{\VV}}$ on $\tilde{\VV} =\wedge^2 \VV$.
 Multiplying this form by the trace zero element $\i$ gives
a {\it skew-hermitian} form on $\tilde{\VV}$, which we denote simply by $(\cdot, \cdot)$.

\begin{lem}
The form $(\cdot, \cdot)$ on $\tilde{\VV}$ satisfies: for all $x,y \in \tilde{\VV}$.
\begin{enumerate}
\item $(x\j , y) = (y\j, x)$.
\item $(x\j,y\j)^\rho = - J (x,y) $.
\end{enumerate}
\end{lem}

\begin{proof} Firstly,
\begin{align*}
(x\j,y) &= \i \cdot [\iota \circ \wedge^2 (\varphi) (x\j) ](y) \\
&= \i \cdot [\psi \circ L \circ L(x)](y) \\
&= J \i \cdot [\psi (x) ](y).
\end{align*}
Since \eqref{eqn:pairing-on-wedge-2} is symmetric, we have
\[
[\psi (x)] (y) = [\psi (y)](x),
\]
from which it follows that $(x\j,y) =(y\j,x)$.

Secondly, we have
\begin{align*}
(x\j, y\j) &=  J \i \cdot [\psi (x) ](y\j) \\
&= J \i \cdot [\psi(y\j)](x) \\
&= J \i \cdot[\psi \circ \psi^{-1} \circ \iota \circ \wedge^2 (\varphi) (y)](x) \\
&= J \cdot (y,x),
\end{align*}
so that
\[
(x\j,y\j)^\rho = J\cdot (y,x)^\rho = -J (x,y).
\]

\end{proof}

\begin{rem}
In fact (ii) above follows from (i). Indeed, assuming (i), we have
\begin{align*}
(x\j, y\j) &= (y\j \cdot \j , x) = J(y,x) = -J(x,y)^\rho.
\end{align*}

\end{rem}

Now we can define a $B$-skew-hermitian form on $\tilde{V}$ by
\begin{equation}
\label{eqn:B-form-defn}
\langle x,y \rangle = (x,y) - \frac{1}{J} \cdot \j \cdot (x\j,y).
\end{equation}

\begin{prop}
The map
\[
\GL_E (\VV) \rightarrow \GL_E(\tilde{\VV}), \quad g \mapsto \wedge^2 g
\]
induces an
isogeny
\[
\GSU_E (\VV) \xrightarrow{\tilde{\xi}} \GU_B (\tilde{V})^0
\]
with kernel $\{ \pm 1 \}$.
\end{prop}

\begin{proof}
Let $g \in \GU_E (\VV)$.
We first compute the commutator of $L$ and $\wedge^2 g$.  Recall that
\[
L= \psi \circ \iota \circ \wedge^2 (\varphi).
\]
Now for any $g\in \GL_E (\VV)$, we have
\begin{equation}
\label{eqn:iota-wedge2g}
\iota \circ \wedge^2 (g) = \wedge^2 (g) \circ \iota
\end{equation}
and
\begin{equation}
\label{eqn:psi-wedge2g}
\psi \circ \wedge^2 g = \det(g) \wedge^2 g \circ \psi.
\end{equation}
If further $g \in \GU_E(\VV)$, then
\[
(gx,gy)_\VV =\nu(g) (x,y)_\VV
\]
which implies that $(gx,y)_\VV = \nu(g) (x, g^{-1} y)_\VV$ and
\[
\varphi \circ g = \nu(g) g \circ \varphi.
\]
Thus
\begin{equation}
\label{eqn:wedge2phi-wedge2g}
\wedge^2 \varphi \circ \wedge^2 g = \nu(g)^2 \wedge^2 g \circ \wedge^2 \varphi.
\end{equation}
It follows from \eqref{eqn:iota-wedge2g}, \eqref{eqn:psi-wedge2g} and \eqref{eqn:wedge2phi-wedge2g} that
\begin{equation}
\label{eqn:L-wedge2g}
L\circ \wedge^2 g=\nu(g)^2 \det(g)^{-1}
 \wedge^2 g \circ L
 \end{equation}
for $g\in \GU_E(\VV)$.
Thus for $g\in \GSU_E(V)$, the endomorphism $\wedge^2(g)$ lies in $\GU_B (\tilde{V})$.
This gives a map
\[
 \GSU_E(V) \rightarrow \GU_B( \tilde{V})
 \]
whose kernel is easily checked to be $\{ \pm 1\}$. From this it follows for
dimension reasons (and since $\GSU_E(V)$ is connected) that the image is the identity component
of $\GU_B(\tilde{V})$.
\end{proof}

\begin{prop}
\label{p:isom-PGU}
There is a natural isomorphism
\[
\xi:  \PGU_E(\VV) \overset{\cong}{\longrightarrow} \PGU_B(\tilde{V})^0,
\]
where
\[
 \PGU_E(\VV) = \GU_E(\VV) / E^\times, \qquad
 \PGU_B(\tilde{V})^0 = \GU_B(\tilde{V})^0 / F^\times.
\]
\end{prop}

\begin{proof}
Let $g \in \GU_E(\VV)$.
Put $f = \wedge^2 g$ and $\alpha = \nu(g)^2 / \det g$.
By \eqref{eqn:L-wedge2g}, we have
\[
 Lf = \alpha fL.
\]
By  Prop. \ref{prop:norm-sim}, we have $\N(\det g) = \nu(g)^4$, so $\N (\alpha)=1$ and we can choose $\beta \in E^\times$ (unique up to multiplication by $F^\times$) such that $\alpha = \beta/\beta^\rho$.
Then
\[
 L \beta f = \beta^\rho L f = \beta^\rho \alpha fL = \beta fL,
\]
so that $\beta f \in \GU_B(\tilde{V})$.
The assignment $g \mapsto \beta f$ gives a homomorphism
\[
 \GU_E(\VV) \longrightarrow \GU_B(\tilde{V})/ F^\times.
\]
It is easy to see that its kernel is the center of $\GU_E(\VV)$ and its image is the identity component of $\GU_B(\tilde{V})/F^\times$.
\end{proof}

\begin{rem}
The reader may note that the notation is mildly confusing here. Namely, $\beta f$ is the map given by $\beta f (x) := f(x) \beta$, since the action of $E$ is on the right.  \end{rem}

\begin{rem}
For $g\in \GSU_E(\VV)$, we have $\alpha=1$, so we may take $\beta=1$ as well. This implies that the maps
constructed in the previous two propositions fit into the commutative diagram below, where the vertical maps
are the natural homomorphisms:
\[
\xymatrix
{
\GSU_E(\VV) \ar[d] \ar[r]^{\tilde{\xi}} & \GU_B(\tilde{V})^0 \ar[dd]\\
\GU_E (\VV)  \ar[d]& \\
\PGU_E(\VV) \ar[r]^{\xi} & \PGU_B (\tilde{V})^0.
}
\]
\end{rem}

\subsection{Subgroups}

In this section we discuss the effect of the isogeny/isomorphism of the previous section on certain natural
subgroups of the unitary group obtained from a decomposition of the hermitian space into
a sum of two hermitian spaces.

\subsubsection{The sum of two $2$-dimensional spaces}
\label{ss:sum-of-2-dim}

We first discuss the case $\VV=\VV_1\oplus \VV_2$, where
$\dim_E \VV_1 = \dim_E \VV_2 = 2$. Then
\begin{equation}
\label{eqn:decomposition-of-VV}
\tilde{\VV} = \wedge^2 (\VV_1 \oplus \VV_2) = \wedge^2 (\VV_1) \oplus \wedge^2 (\VV_2) \oplus (\VV_1 \otimes \VV_2),
\end{equation}
where we identify $\VV_1 \otimes \VV_2$ with its image under the natural map to $\wedge^2 \VV$ (which sends
$v\otimes w$ to $v\wedge w$.)

We may assume that the basis $(\v_1, \v_2, \v_3, \v_4)$ of
$\VV$ is chosen such that $(\v_1,\v_2)$ forms a basis of $\VV_1$
and $(\v_3, \v_4)$ a basis of $\VV_2$. From the explicit formulas for $L$, it is clear that $L$ preserves the subspaces
\[
\VV_0^\sharp= \wedge^2 (\VV_1) \oplus \wedge^2 (\VV_2) \quad \text{ and} \quad \VV^\sharp= \VV_1 \otimes \VV_2,
\]
so these are $B$-spaces that we denote by $V_0^\sharp$ and $V^\sharp$ respectively.
Since the collection $(\v_{ij}$, $1\le i < j \le 4)$ forms an orthogonal basis for the form $(\cdot,\cdot)$, the
decomposition $\tilde{\VV} = \VV^\sharp \oplus \VV_0^\sharp$ is one of skew-hermitian $E$-spaces.  Moreover, the formula \eqref{eqn:B-form-defn} shows that
the decomposition $\tilde{V} = V^\sharp \oplus V_0^\sharp$ is one of skew-hermitian $B$-spaces.

\begin{prop}
Let $H$ be the subgroup of $\GSU_E (\VV)$ given by
\[
H= \GSU_E (\VV) \cap \G (\U_E (\VV_1) \times \U_E (\VV_2)).
\]
Then $\tilde{\xi}$ restricts to an isogeny
\begin{equation}
\label{eqn:xit-rest}
H \rightarrow \G(\U_B(V^\sharp) \times \U_B (V_0^\sharp))^0,
\end{equation}
with kernel $\{ \pm 1 \}$.
\end{prop}

\begin{proof}
Let $g_1 \in \GU_E(\VV_1)$, $g_2 \in \GU_E( \VV_2)$ be such that
$g=(g_1,g_2) \in H$. Then $\wedge^2 g$ acts as right multiplication by $\det(g_i)$ on
$\wedge^2 \VV_i$ and by $g_1 \otimes g_2$ on $\VV_1 \otimes \VV_2$. This shows that
$\tilde{\xi}$ maps $H$ into the subgroup $ \G(\U_B(V^\sharp) \times \U_B (V_0^\sharp))^0$ of
$\GU_B (\tilde{V})^0$. By counting dimensions, we see that the restriction of
$\tilde{\xi}$ is an isogeny with kernel $\{ \pm 1 \}$. \end{proof}

Likewise, one has an analogous result for subgroups in the context of Prop. \ref{p:isom-PGU}.

\begin{prop}
The map $\xi$ restricts to give an isomorphism
\begin{equation}
\label{eqn:rest-of-xi}
\G (\U_E(\VV_1) \times \U_E (\VV_2)) / E^\times \simeq \G (\U_B( V^\sharp) \times \U_B (V_0^\sharp))^0/F^\times.
\end{equation}
\end{prop}

\begin{proof}
Let $g=(g_1,g_2) \in \G (\U_E(\VV_1) \times \U_E (\VV_2))$. As in the previous proposition, the map $\wedge^2 g$ clearly preserves the decomposition
$\tilde{\VV} = \VV^\sharp \oplus \VV_0^\sharp$, hence so does $\beta \cdot \wedge^2 g$. Since $\beta \cdot \wedge^2 g$ lies in $\GU_B (\tilde{V})$,
it must in fact lie in $\G (\U_B( V^\sharp ) \times \U_B (V_0^\sharp))$. This gives the map \eqref{eqn:rest-of-xi}, which must be injective since
$\xi$ is injective. From dimension considerations, it must be an isomorphism.
\end{proof}

It is useful to write down explicitly the maps in \eqref{eqn:xit-rest} and \eqref{eqn:rest-of-xi}. First for
$i=1,2$, we define a quaternion algebra $B_i$ such that
$\VV_i$ is naturally a right-$B_i$ module and is equipped with a
$B_i$-hermitian form whose projection to $E$ recovers the hermitian form.
Let us outline for $i=1$, the case $i=2$ being exactly similar. The hermitian form on
$\VV_1$ gives an $E$-conjugate linear isomorphism
\[
\varphi_1: \VV_1 \simeq \VV_1^*,
\]
where $\VV_1^*$ as usual is the $E$-linear dual of $\VV_1$. Let us fix an isomorphism
\[
d_1: \wedge^2 \VV_1 \simeq E.
\]
For definiteness, we let $d_1 (\v_1 \wedge \v_2) = 1$. This gives a bilinear pairing
\[
\VV_1 \times \VV_1 \rightarrow E, \quad (x,y) \mapsto d_1 (x\wedge y)
\]
and thus gives an $E$-linear isomorphism
\[
\psi_1: \VV_1 \simeq \VV_1^*, \quad \psi_1 (x) (y) = d_1 (x\wedge y).
\]
Define $L_1 \in \End_F(\VV_1)$ by
\[
L_1 = \psi_1^{-1} \circ \varphi_1.
\]
Explicitly, we see that $L_1$ acts on $\VV_1$ by:
\begin{align}
\label{eqn:L11} \v_1 & \mapsto -a_1 \v_2 \\
\label{eqn:L12} \v_2 & \mapsto a_2 \v_1,
\end{align}
so that $L_1^2 = -a_1a_2 = -\vol (\VV_1, d_1)$.
Let
\[
J_1 = -\vol(\VV_1, d_1)
\]
and define $B_1$ by
\[
B_1 = E+ E\j_1, \quad \j_1^2 = J_1, \quad \alpha \j_1 = \j_1 \alpha^\rho
\]
for all $\alpha \in E$. Then the right $E$-action on $\VV_1$ extends to a right $B_1$-action
defined by
\[
x \cdot \j_1 = L_1 (x).
\]
When we want to think of $\VV_1$ as a $B_1$-space we simply denote it $V_1$.

\begin{lem}
For $x,y \in \VV_1$, we have
\begin{enumerate}
\item $(x\j_1, y) = - (y\j_1, x)$.
\item $(x\j_1, y\j_1)^\rho = -J(x,y) $.
\end{enumerate}
\end{lem}

\begin{proof}
We have for $x,y \in \VV_1$,
\[
(x\j_1, y) = \varphi_1 (x\j_1)  (y) = \psi_1 L_1 (x\j_1) (y) = \psi_1 L_1^2 (x) (y) = J_1 \psi_1 (x) (y) =J_1 d_1 (x\wedge y),
\]
and likewise $(y\j_1, x) = J_1 d_1 (y \wedge x)$. It follows from this that $(x\j_1, y) = - (y\j_1,x)$ which proves (i). Now (ii)
follows from (i) since
\[
(x\j_1, y\j_1) = -(y\j_1 \cdot \j_1, x) = -J_1(y,x) = -J_1 (x,y)^\rho.
\]
\end{proof}

Using the lemma, we find that
\[
\langle x,y \rangle := (x,y) - \frac{1}{J_1} \j_1 (x\j_1, y)
\]
defines a $B$-{\it hermitian} form on $V_1$ such that $\pr \circ \langle \cdot,\cdot \rangle = (\cdot, \cdot)$. Thus there
is a natural embedding
\[
\GU_{B_1}(V_1) \hookrightarrow \GU_E(\VV_1)
\]
which we can make explicit as follows. Pick a $B_1$-basis $\x_1$ for $V_1$. Then
$V_1 = \x_1 B_1$ and for $g\in \GU_{B_1}(V_1)$, let $\beta_g$ be defined by
\[
g\x_1 = \x_1 \beta_g.
\]
The assignment $g\mapsto \beta_g$ gives an identification $\GU_{B_1} (V_1) \simeq B_1^\times$.
Indeed, this map is injective; it is also surjective since for any $\beta\in B_1^\times$ and $\alpha,\alpha' \in B_1$, we have
\[
\langle \x_1 \beta \alpha ,\x_1 \beta \alpha' \rangle = (\beta\alpha)^\rho \langle \x_1,\x_1 \rangle \beta \alpha' = \alpha^\rho \beta^\rho \langle \x_1,\x_1 \rangle   \beta \alpha' = \nu (\beta)  \alpha^\rho \langle \x_1,\x_1 \rangle
\alpha' = \nu(\beta) \langle \x_1 \alpha, \x_1 \alpha' \rangle.\]
Here we have used that $\langle \x_1,\x_1 \rangle$ lies in $F$, the form $\langle \cdot, \cdot \rangle$ being $B$-hermitian.
Then
\[
\GU_E (\VV_1) \simeq (B_1^\times \times E^\times)/F^\times
\]
where the $B_1^\times$ corresponds to the $\GU_{B_1} (V_1)$ action described above, the $E^\times$-action is given by $\alpha \mapsto $ (right)-multiplication by $\alpha^{-1}$ for $\alpha \in E^\times$ and the embedding of
$F^\times$ in $B_1^\times \times E^\times$ is just the diagonal embedding $t\mapsto (t,t)$. All of the above discussion carries over verbatim to the case $i=2$, so that after picking a
$B_2$-basis $\x_2$ of $V_2$, the
map $\GU_{B_2} (V_2) \hookrightarrow \GU_E (\VV_2)$ is identified with the embedding
\[
B_2^\times \hookrightarrow (B_2^\times \times E^\times)/F^\times.
\]

Next, we explicate the groups on the right of the map \eqref{eqn:xit-rest}.
First we note that
\[
\GU_B(V_0^\sharp)^0 \simeq E^\times.
\]
Indeed, let $\x$ be any non-zero vector in $\wedge^2 V_1$, so that $V_0^\sharp =\x B$. Then for $\alpha \in E^\times$, the map
\[
\x \beta \mapsto \x \alpha \beta
\]
gives an element of $\GU_B (V_0^\sharp)$. By dimension considerations, this gives an isomorphism $E^\times \simeq \GU_B(V_0^\sharp)^0$.
More precisely, it is easy to see that for any $g\in \GU_B (V_0^\sharp)$, we have $g \cdot \x = \x \gamma_g$ for a unique $\gamma_g \in B^\times$ and the assignment
$g\mapsto \gamma_g$ gives an isomorphism of $\GU_B( V_0^\sharp)$ with the semi-direct product
$E^\times \rtimes  \langle \j \rangle$,
where $\j$ acts on $E^\times$ by conjugation in $B^\times$. Note that these isomorphisms are independent of the choice of $\x \in \wedge^2 V_1$.

As for $V^\sharp$, we have the following proposition.
\begin{prop}
There is an isomorphism
\[
\GU_B (V^\sharp)^0 \simeq (B_1^\times \times B_2^\times)/F^\times,
\]
depending on the choice of a basis vector for $V_1$ (as $B_1$-space) and for $V_2$ (as $B_2$-space).
\end{prop}

\begin{proof}
First note that the restriction of the hermitian form $(\cdot, \cdot)_{\tilde{\VV}}$ to
$\VV_1 \otimes \VV_2$ is just the tensor product of the hermitian forms on $\VV_1$ and $\VV_2$.
Also, from \eqref{eqn:L11} and \eqref{eqn:L12}, right multiplication by $\j_1$ and $\j_2$ on
$\VV_1$ and $\VV_2$ is given explicitly by
\[
 \v_1 \j_1 = -a_1 \v_2, \quad  \v_2 \j_1 = a_2 \v_1, \quad  \v_3 \j_2 = -a_3 \v_4, \quad
 \v_4 \j_2 = a_4 \v_3.
\]
From this and the explicit formula for the action of $L=\j$ on $\tilde{\VV}$, we see that
(right)-multiplication by $\j$ on $V_1 \otimes V_2$ is the same as (right)-multiplication by
$\j_1 \otimes \j_2$.

Choose a $B_1$-basis $\x_1$ for $V_1$ and a $B_2$-basis $\x_2$ for $V_2$.
Then there is an action of $B_1^\times \times B_2^\times$ on $\VV_1 \otimes_E \VV_2$ which
on pure tensors is given by
\[
(\beta_1, \beta_2) \cdot (\x_1 \alpha_1 \otimes \x_2 \alpha_2) = \x_1 \beta_1 \alpha_1 \otimes \x_2 \beta_2 \alpha_2,
\]
for any $\alpha_1 \in B_1$, $\alpha_2 \in B_2$. This action is clearly $E$-linear and also $\j$-linear since $\j$ acts
as right multiplication by $\j_1 \otimes \j_2$, hence is in fact $B$-linear.

Now,
\begin{align*}
(\x_1 \beta_1 \alpha_1 \otimes \x_2 \beta_2 \alpha_2, \x_1 \beta_1 \alpha'_1 \otimes \x_2 \beta_2 \alpha'_2) &= \i \cdot (\x_1 \beta_1 \alpha_1 ,  \x_1 \beta_1 \alpha'_1)_{\VV_1} (\x_2 \beta_2 \alpha_2,\x_2 \beta_2 \alpha'_2)_{\VV_2} \\
&= \i \cdot \nu(\beta_1) \nu (\beta_2) \cdot (\x_1  \alpha_1 ,  \x_1  \alpha'_1)_{\VV_1} (\x_2 \alpha_2,\x_2  \alpha'_2)_{\VV_2} \\
&= \nu(\beta_1) \nu (\beta_2) (\x_1 \alpha_1 \otimes \x_2 \alpha_2, \x_1 \alpha'_1 \otimes \x_2 \alpha'_2),
\end{align*}
which shows that $(\beta_1, \beta_2)$ gives an element in $\GU_E (\VV^\sharp)$. Since the action of
$(\beta_1, \beta_2)$ commutes with $\j$, we see from the formula \eqref{eqn:B-form-defn} that
it in fact defines an element of $\GU_B (V^\sharp)$. This gives a map
\[
B_1^\times \times B_2^\times \rightarrow \GU_B(V^\sharp)
\]
whose kernel is the diagonal $F^\times $ in $B_1^\times \times B_2^\times$, embedded as
$t \mapsto (t,t^{-1})$. By dimension considerations, we see that this gives an isomorphism
$(B_1^\times \times B_2^\times)/F^\times \simeq \GU_B( V^\sharp)^0$.
\end{proof}

Now we can write down the map \eqref{eqn:xit-rest} explicitly. Let $h=(h_1,h_2) \in H$ with $h_1 \in \GU_E (\VV_1)$ and $h_2 \in \GU_E (\VV_2)$.
Then
\[
\nu(h_1) = \nu(h_2)
\]
and
\[
\det(h_1) \det(h_2) = \nu(h_1)^2 = \nu (h_2)^2.
\]
Moreover, $\N (\det h_i) = \nu (h_i)^2$ by Prop. \ref{prop:norm-sim}, so that $\det(h_1) = \det(h_2)^\rho$.
Now $\wedge^2 h$ acts as (right)-multiplication by $\det(h_1)$ on $\wedge^2 \VV_1$ and by $\det(h_2)$ on $\wedge^2 \VV_2$,
and thus acts on $V_0^\sharp$ as the element $\det(h_1) \in E^\times = \GU_B(V_0^\sharp)^0$.

Fix a basis vector $\x_1$ for $V_1$ and $\x_2$ for $V_2$ as above. This gives identifications
\[
\GU_{B_1} (V_1) = (B_1^\times \times E^\times)/F^\times, \quad \GU_{B_2} (V_2) = (B_2^\times \times E^\times)/F^\times.
\]
Let $g_1=[(b_1,\alpha_1)] \in  \GU_E (\VV_1)$ and $g_2=[(b_2,\alpha_2)] \in  \GU_E (\VV_2)$. Then
\[
\nu(g_1) = \nu(b_1) \N(\alpha_1)^{-1},  \quad \det(g_1)= \nu(b_1) \cdot \alpha_1^{-2},
\]
and
\[
\nu(g_2) = \nu(b_2) \N(\alpha_2)^{-1},  \quad \det(g_2)= \nu(b_2) \cdot \alpha_2^{-2}.
\]
Let $g=(g_1, g_2) \in \G (\U_E (\VV_1) \times \U_E (\VV_2))$ viewed as an element in $\GU_E(\VV)$.
Then
\[
\nu(g) = \nu(g_1) =\nu(g_2) = \nu(b_1) \N(\alpha_1)^{-1} = \nu(b_2) \N(\alpha_2)^{-1}
\]
while
\[
\det (g) =  \nu(b_1) \cdot \alpha_1^{-2} \cdot  \nu(b_2) \cdot \alpha_2^{-2}.
\]
Thus
\begin{align*}
g \in \GSU_E(\VV) & \iff \det(g) = \nu(g)^2 \\
& \iff  \nu(b_1) \cdot \alpha_1^{-2} \cdot  \nu(b_2) \cdot \alpha_2^{-2} = \nu(b_1) \N(\alpha_1)^{-1} \cdot \nu(b_2) \N(\alpha_2)^{-1} \\
& \iff \alpha_1 \alpha_2 = \alpha_1^\rho \alpha_2^\rho \\
& \iff \alpha_1 \alpha_2 \in F^\times.
\end{align*}
We conclude that
\[
H = \left\{ \left( [(b_1,\alpha_1)], [(b_2,\alpha_2] \right):  \nu(b_1) \N(\alpha_1)^{-1} = \nu(b_2) \N (\alpha_2)^{-1} , \ \alpha_1 \alpha_2 \in F^\times \right\}
\]
The action of $h= \left( [(b_1,\alpha_1)], [(b_2,\alpha_2] \right)$ on $\VV_1 \otimes_E \VV_2$ is then given by
\[
\x_1 \beta_1 \otimes \x_2 \beta_2 \mapsto \x_1 b_1 \beta_1 \alpha_1^{-1} \otimes \x_2 b_2 \beta_2 \alpha_2^{-1} = \x_1 b_1 \beta_1 \alpha_1^{-1} \alpha_2^{-1} \otimes \x_2 b_2 \beta_2 =\x_1 b_1 (\alpha_1 \alpha_2)^{-1} \beta_1 \otimes \x_2 b_2 \beta_2,
\]
so that the map \eqref{eqn:xit-rest} is given explicitly by:
\[
h \mapsto  \left([(b_1 (\alpha_1 \alpha_2)^{-1}, b_2)],  \nu(b_1) \alpha_1^{-2} \right) \in \G ( (B_1^\times \times B_2^\times)/F^\times) \times E^\times).
\]

Likewise, we can make \eqref{eqn:rest-of-xi} explicit. Let $g =(g_1,g_2) \in \G (\U_E(\VV_1) \times \U_E (\VV_2)) $, with $g_1 =[(b_1,\alpha_1)]$ and $g_2=[(b_2,\alpha_2)]$.
Then
\[
\frac{\nu(g)^2}{\det (g)} = \frac{\nu (b_1) \N (\alpha_1)^{-1} \nu(b_2) \N (\alpha_2)^{-1} }{\nu(b_1) \nu(b_2) \alpha_1^{-2} \alpha_2^{-2}} = \frac{\alpha_1 \alpha_2}{(\alpha_1\alpha_2)^\rho},
\]
so we may take $\beta=\alpha_1 \alpha_2$ in the definition of $\xi(g)$. Then the action of $\xi$ on $g$ is given by
\[
g \mapsto ([(b_1, b_2)], \det(g_1) (\alpha_1 \alpha_2)).
\]

\begin{example}
This is the case that is of most interest to us. Instead of starting with the spaces
$\VV$ or $\VV_1$ or $\VV_2$, we start with two quaternion algebras
$B_1$ and $B_2$ over $F$ containing $E$. Suppose that
\[
B_1 = E + E \j_1, \quad B_2 = E+ E \j_2,
\]
with $\j_1^2 = J_1$ and $\j_2^2 = J_2$.
Let $\VV_i = B_i$ considered as a (right)-$E$-hermitian space with the same form as in periods1, i.e.,
\[
(a+\j_i b,c+\j_i d)_i = a^\rho c - J_i b^\rho d.
\]
We specialize the setup above to the case:
\[
\VV = \VV_1 \oplus \VV_2,
\]
with hermitian form $(\cdot, \cdot)_\VV$ given by the direct sum of $(\cdot, \cdot)_1$ and $(\cdot,\cdot)_2$.
In the basis
\[
\v_1=(1,0), \  \v_2=(\j_1,0), \ \v_3=(0,1), \ \v_4=(0,\j_2),
\]
this form is diagonal with matrix
\[
\begin{bmatrix} a_1 & & & \\ & a_2 & & \\ & & a_3 & \\ & & & a_4 \end{bmatrix}  =
\begin{bmatrix} 1 & & & \\ & -J_1 & & \\ & & 1 & \\ & & & -J_2 \end{bmatrix}.
\]

Let us pick $d:\wedge^4 \VV \simeq E$ such that
\[
d (\v_1 \wedge \v_2 \wedge \v_3 \wedge \v_4) = -1
\]
as in the proof of Prop. \ref{prop:volume-V-d}.

With respect to the bases $(\v_1 \wedge \v_2, \v_3 \wedge \v_4 )$ of  $\VV_0^\sharp$ and
$(\v_1 \otimes \v_3, \v_2 \otimes \v_3 , \v_1 \otimes \v_4, \v_2 \otimes \v_4)$ of $\VV^\sharp$, the
matrices of these hermitian forms are given by
\[
\begin{bmatrix} -J_1 & \\ & - J_2 \end{bmatrix}, \quad \text{and} \quad
\begin{bmatrix} 1 & & & \\ & -J_1 & & \\ & & -J_2 & \\ & & & J \end{bmatrix}
\]
respectively. Thus $\VV^\sharp$ is the tensor product of
$\VV_1$ and $\VV_2$ as {\it hermitian} spaces. The action of
$\j$ on $\VV^\sharp$ can be read off from the formulas in the proof of Prop. \ref{prop:volume-V-d} and is given by:
\begin{align*}
(\v_1 \otimes \v_3)\cdot \j &= \v_2 \otimes \v_4, \\
(\v_2 \otimes \v_3) \cdot \j &= J_1 \cdot \v_1 \otimes \v_4, \\
(\v_1 \otimes \v_4) \cdot \j & = J_2 \cdot \v_2 \otimes \v_3, \\
(\v_2 \otimes \v_4) \cdot \j &= J \cdot \v_1 \otimes \v_3.
\end{align*}
This shows that $V^\sharp$ with its $B$-action and $B$-skew-hermitian form is
exactly the same as the space $V$ occurring in periods1.
\end{example}

\subsubsection{The sum of a $3$-dimensional and a $1$-dimensional space}

In this section (which will only be used in a later paper) we suppose that $\VV= \VV_3 \oplus \VV_4$, where $\dim \VV_3 = 3$ while $\dim \VV_4 =1$.
Then there is an inclusion
\[
\G(\U_E(\VV_3) \times \U_E(\VV_4)) \hookrightarrow \GU_E (\VV)
\]
and so we can ask for a description of how this relates to the maps $\tilde{\xi}$, $\xi$.

Note that
\[
\wedge^2 \VV = \wedge^2 \VV_3 \oplus (\VV_3 \otimes \VV_4)
\]
as a sum of (skew)-hermitian spaces.
We may assume that the basis vectors $\v_i$ are chosen such that
$(\v_1, \v_2, \v_3)$ forms a basis for $\VV_3$ while $\v_4$ is a basis for $V_4$.
The explicit formula for $L$ shows that $L$ interchanges $\wedge^2 \VV_3$ and
$\VV_3 \otimes \VV_4$. Thus letting $\WW:=\VV_3 \otimes \VV_4$, we have
\begin{equation}
\label{eqn:extendingW}
\tilde{V} = \WW \otimes _E B,
\end{equation}
at least as $B$-spaces. The formula \eqref{eqn:B-form-defn} shows that the
restriction of the  $B$-skew-hermitian form $\langle \cdot, \cdot \rangle$ to $\WW$ is the same as
the restriction of the $E$-skew-hermitian $(\cdot, \cdot)$ (from $\tilde{\VV}$) to $\WW$, from which it follows that
the form $\langle \cdot, \cdot \rangle$ on $\tilde{V}$ is just the $B$-linear extension of $(\cdot ,\cdot)$ via the isomorphism
\eqref{eqn:extendingW}. Thus there is a canonical inclusion
\[
\GU_E(\WW) \rightarrow \GU_B (\tilde{V}),
\]
which must land inside $\GU_B(\tilde{V})^0$, since $\GU_E(\WW)$ is connected.

\begin{prop}
The map $\xi$ restricts to an isomorphism
\[
\G (\U_E (\VV_3) \times \U_E (\VV_4))/E^\times \simeq \GU_E(\WW)/F^\times.
\]
\end{prop}

\begin{proof}
It is clear that $\xi$ restricts to an injective map between the given source and target. That it is an isomorphism follows from dimension considerations and since the target is connected.
\end{proof}

config: arxiv


TITLE: Evaluating using Contour Integration
QUESTION [0 upvotes]: Another revision question I need a bit of help with, sorry guys!
By choosing an appropriate contour, evaluate the following integral;
$$\int_{-\infty}^\infty \frac{e^{iz}}{(z-i)^2}$$
Every time I attempt this question, I get the denominator to equal 0 somehow. I'm sure there must be a smarter way to manipulate the denominator into getting the z not equal to i, but I'm stuck how to at the moment. Thanks in advance.

REPLY [3 votes]: For a large value of $R\in\mathbb{R}^+$, consider the countour $\gamma_R$ made by a segment $S_R$ between $-R$ and $R$ and the semicircle $\Gamma_R$ in the upper half-plane joining $R$ with $-R$ (let all the path be counter-clockwise oriented). By the residue theorem,
$$\oint_{\gamma_R}\frac{e^{iz}}{(z-i)^2}\,dz = 2\pi i\cdot\text{Res}\left(\frac{e^{iz}}{(z-i)^2},z=i\right)=2\pi i\cdot \frac{i}{e}=\color{red}{-\frac{2\pi}{e}}$$
and you just need to prove that
$$ \lim_{R\to +\infty}\int_{\Gamma_R}\frac{e^{iz}}{(z-i)^2}\,dz = 0.$$