Calculating the Galois cohomology of $U(n)$

Question

I would like to know know what the Galois cohomology $H^1(Gal( \Bbb{C}/\Bbb{R}),U(n)(\Bbb{C}) ) $ is.

The unitary group $U(n)$ can be defined as the real form of $Gl_{n,\Bbb{C}}$ relative to the involution $\phi$ sending $A\mapsto (\bar{A}^\top)^{-1}$. Then explicitily $$H^1(Gal( \Bbb{C}/\Bbb{R}),U(n)(\Bbb{C}) )=\{g\in GL( \Bbb{C}^n)\mid g \phi(g)=I\} /\sim $$ where the equivalence relation is induced by $ g\sim Ag\phi(A)^{-1}$ for $A\in GL(\Bbb{C}^n)$. See here for the exact definition of the first non-abelian cohomology group.

The case $n=1$ is easy and shows that the Galois cohomology is trivial, but I assume it is non-trivial for higher $n$.

Answer 1

$\newcommand{\N}{\mathbb{N}}$$\newcommand{\Res}{\mathrm{Res}}$$\newcommand{\ad}{\mathrm{ad}}$$\newcommand{\bb}[1]{\mathbb{#1}}$$\newcommand{\C}{\mathbb{C}}$$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Gal}{\mathrm{Gal}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\GL}{\mathrm{GL}}$

Notation and terminology

To make our lives easier, let us fix the following notation/terminology:

• if $G$ is an algebraic group over a field $F$ then we shall write $H^i(F,G)$ as shorthand for $H^i(\Gal(\ov{F}/F),G(\ov{F})$,
• for such a $G$, let us also write $G^\ast$ for the unique quasi-split inner form of $G$ (e.g., see [Mil, Corollary 23.54]),
• a form that is not inner is called outer.

Forms of $\mathrm{GL}_{n,\mathbb{R}}$

Let us begin by clarifying that $U(n)$ is ambiguous. Over $\R$ there are many groups which are called 'unitary groups'. I think the following is probably the least contentious definition.

Definition 1: A unitary group over $\mathbb{R}$ is an outer form of $\mathrm{GL}_{n,\mathbb{R}}$.

Now, the forms of $\mathrm{GL}_{n,\mathbb{R}}$ are easy to understand. There is a decomposition

$$\mathrm{Form}(\mathrm{GL}_{n,\mathbb{R}})=\bigsqcup_{x\in X}\mathrm{InnForm}(G_x),$$

where $X$ is a set of representatives from the inner-form classes of the forms of $\mathrm{GL}_{n,\mathbb{R}}$, and $G_x$ is a(ny) representative of $X$. But, we can easily compute $\#(X)$. Indeed, as in [Mil, Theorem 23.52] there is a bijection

$$X\xrightarrow{\sim} H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),\mathrm{Out}(\mathrm{GL}_{n,\mathbb{C}})),$$

where $\mathrm{Out}(\mathrm{GL}_{n,\mathbb{C}})=\mathrm{Aut}(\mathrm{GL}_{n,\mathbb{C}})/\mathrm{Inn}(\mathrm{GL}_{n,\mathbb{R}})$ has the obvious Galois action. (I'm being a little imprecise here about outer forms vs. outer twists, but let's not worry about that)

But, the outermorphism group of $\mathrm{GL}_{n,\mathbb{R}}$ is easy to compute using root data (e.g., see [Mil, Corollary 23.46]) and in this case it's easy to compute that it's $\mathbb{Z}/2$ with the trivial Galois action (the non-trivial outermorphism corresponding to $g\mapsto (g^t)^{-1}$). So, it's then easy to see that

$$\#(X)=\# H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),\mathrm{Out}(\mathrm{GL}_{n,\mathbb{C}}))=\# \mathrm{Hom}(\mathbb{Z}/2,\mathbb{Z}/2)=2.$$

As we may certainly take $\mathrm{GL}_{n,\mathbb{R}}$ as one element of $X$, it suffices to find any other outer form $G$ of $\mathrm{GL}_{n,\mathbb{R}}$ to fill out $X$.

To this end, for $p,q\in\N$ such that $p+q=n$ let us consider

$$U(p,q)=\left\{g\in \Res_{\C/\R}\,\GL_{n,\C}: g^\ast J_{p,q} g=J_{p,q}\right\},$$

where $g^\ast=\ov{g}^t$, and $J_{p,q}=\mathrm{antidiag}(I_p,-I_q)$, where $I_m$ is the $m\times m$ identity matrix.

The following lemma is elementary and left to the reader.

Lemma 2: For any $p,q\in\N$ with $p+q=n$ the group $U(p,q)\simeq U(q,p)$. Moreover, if $p\geqslant q$, then $U(p,q)$ is a form of $\mathrm{GL}_{n,\mathbb{R}}$, has maximally split maximal torus isomorphic to $(\Res_{\C/\R}\,\mathbb{G}_{m,\mathbb{C}})^{q}\times U(1)^{p-q}$, and center isomorphic to $U(1)$.

Here

$$U(1)=\left\{\begin{pmatrix}a& b\\ -b & a\end{pmatrix}\in \Res_{\C/\R}\,\GL_{n,\R}: a^2+b^2=1\right\},$$

is the circle group, a (the) non-split one-dimensional torus over $\mathbb{R}$.

From this, it's easy to deduce that among pairs $p,q\in\N$ with $p+q=n$ and $p\geqslant q$ that

• $U(p,q)$ is an outer form of $\mathrm{GL}_{n,\R}$ (as inner forms have isomorphic centers),
• no two are isomorphic to each other (by counting the split rank of a maximally split maximal torus).

Thus, any $U(p,q)$ may constitute the other element of $X$ and, in fact, we thus deduce that any two $U(p,q)$ are an inner forms of each other.

Let us write $U(n):=U(n,0)\simeq U(0,n)$. Then, we conclude that

$$\mathrm{Form}(\mathrm{GL}_{n,\mathbb{R}})=\mathrm{InnForm}(\mathrm{GL}_{n,\mathbb{R}})\sqcup \mathrm{InnForm}(U(n))\qquad (1).$$

Galois cohomology of $U(n)$

Let us begin by calculating the Galois cohomology groups $H^1(\mathbb{R},U(n))$ and $H^1(\mathbb{R},U(n)^\mathrm{ad})$.

Our key result is the following:

Theorem 3([Bor, Theorem 3.1]): Let $G$ be a reductive group over $\mathbb{R}$ and let $T$ be a fundamental torus of $G$ (i.e., one of minimal split rank). Then, there is a canonical bijection $$H^1(\mathbb{R},T)/W_T(\mathbb{R})\to H^1(\mathbb{R},G).$$

Here $W_T$ be the Weyl group scheme, $$W_T:=N_G(T)/T.$$ Observe that $W_T(\mathbb{R})$ naturally acts on $H^1(\mathbb{R},T)$ in way equivariant for for the projection to $H^1(\mathbb{R},G)$, so this induces the natural map in the theorem.

Let us apply this to the case when $U(n)$. In particular, note that by Lemma 2, $T=U(1)^n$ is a fundamental torus of $U(n)$. So, we first need to calculate $H^1(\mathbb{R},U(1))$.

Lemma 4: The following equalities holds for $i>0$:

$$H^i(\R,U(1))=\begin{cases}\R^\times/\mathrm{Nm}(\C^\times) & \mbox{if}\quad i\text{ is odd}\\ 0 & \mbox{if}\quad i\text{ is even}.\end{cases}$$ $$H^i(\R,\mathbb{G}_{m,\mathbb{R}})=\begin{cases}\R^\times/\mathrm{Nm}(\C^\times) & \mbox{if}\quad i\text{ is even}\\ 0 & \mbox{if}\quad i\text{ is odd}.\end{cases},$$ where $\mathrm{Nm}\colon \C^\times\to\R^\times$ is the norm map.

Proof: We have a short exact sequence

$$1\to U(1)\to \Res_{\C/\R}\,\bb{G}_{m,\C}\xrightarrow{\mathrm{Nm}} \bb{G}_{m,\R}\to 1,$$

where $\mathrm{Nm}$ is the norm map. We then get a long exact sequence

$$1\to U(1)(\R)\to \C^\times\to\R^\times\to H^1(\R,U(1))\to H^1(\R,\Res_{\C/\R}\,\bb{G}_{m,\C})\to H^1(\R,\bb{G}_{m,\R})\to\cdots.$$

By Shapiro's lemma we know that

$$H^i(\R,\Res_{\C/\R}\bb{G}_{m,\C})=0,\qquad i>0$$

And, thus we deduce that $$H^{i+1}(\R,U(1))\simeq H^{i}(\R,\mathbb{G}_{m,\R})$$ for $i>1$. Moreover, we see that $$H^1(\R,U(1))=\C^\times/\mathrm{Nm}(\R^\times).$$ We thus reduce the calculation for $U(1)$ to that of $\mathbb{G}_{m,\mathbb{R}}$ which is easy because one is just computing $H^i(\mathbb{Z}/2,\mathbb{C}^\times)$ where the non-trivial element of $\Z/2$ acts by conjugation. $\blacksquare$

Of course, $\mathrm{Nm}(\C^\times)=\R^{>0}$ and thus $\R^\times/\mathrm{Nm}(\C^\times)\cong\Z/2$.

$$H^1(\mathbb{R},T)=H^1(\mathbb{R},U(1)^n)=H^1(\mathbb{R},U(1))^n=(\mathbb{Z}/2)^n$$

To proceed further we ned the following fun exercise (a proof can be found at loc. cit.).

Lemma 5: Let $G$ be a reductive group over $\mathbb{R}$ and suppose that $T$ is a fundamental torus. Then, the group scheme $W_T$ is constant.

In particular, returning to the case of $U(n)$ and $T=U(1)^n$, we see that $$W_T(\mathbb{R})=W_T(\mathbb{C})=S_n,$$ where we are using the fact that $U(n)_{\mathbb{C}}\simeq \mathrm{GL}_{n,\mathbb{C}}$ whose Weyl group is $S_n$, acting by permuting the factors. In particular, It's not hard to see that $S_n$ acts on

$$H^1(\mathbb{R},U(1)^n)\simeq H^1(\mathbb{R},U(1))^n\simeq (\C^\times/\mathrm{Nm}(\R^\times))^n\simeq (\mathbb{Z}/2)^n,$$ by permuting factors the factors.

Thus, making the identification

$$(\mathbb{Z}/2)^n/S_n\simeq\{(p,q)\in\mathbb{N}^2:p+q=n\},$$

we deduce the following calculation.

Proposition 6: There is a natural bijection $$H^1(\mathbb{R},U(n))\simeq \{(p,q)\in\mathbb{N}^2:p+q=n\}.$$

To compute $H^1(\mathbb{R},U(n)^\mathrm{ad})$ we could repeat a similar argument. But, perhaps it's fun to instead use the exact sequence

$$H^1(\R,U(1))\to H^1(\R,U(n))\to H^1(\R,U(n)^{\ad})\to H^2(\R,U(1)).$$

By Lemma 4 this last group vanishes and the first group is isomorphic to $\Z/2$, and so we arrive at the exact sequence. Thus, we arrive at the sequence

$$\Z/2\Z\to H^1(\R,U(n))\to H^1(\R,U(n)^{\ad})\to 0.$$

By [Ser,§5.7], we deduce that there is a natural $\Z/2\Z$ action on $H^1(\R,U(n))$ whose orbit space can be identified with $H^1(\R,U(n)^{\ad}$. One can verify that the non-trivial element of $\Z/2$ acts on

$$H^1(\mathbb{R},U(n))\simeq \{(p,q)\in\mathbb{N}^2:p+q=n\},$$

is via the transposition $(p,q)\mapsto (q,p)$.

Thus, all in all, we deduce the following

Proposition 7: The map $$H^1(\mathbb{R},U(n))\to H^1(\R,U(n)^\mathrm{ad})$$ is surjective and may be identified with the obvious map $$\{(p,q)\in\mathbb{N}^2:p+q=n\}\to \{(p,q)\in\mathbb{N}^2:p+q=n\text{ and }p\geqslant q\}.$$

Forms of $\mathrm{GL}_{n,\mathbb{R}}$

We can now actually completely describe all of the forms of $\mathrm{GL}_{n,\mathbb{R}}$ and their Galois cohomology. By Equation (1) it suffices to compute the inner forms of $\mathrm{GL}_{n,\mathbb{R}}$ and $U(n)$. We then use the following fact.

Lemma 8 (cf., [Mil, Corollary 23.48]): Let $G$ be a reductive group over $\mathbb{R}$. There is a bijection $$\mathrm{InnForm}(G)\simeq H^1(\R,G^\ad).$$

Now, in Proposition 7 we already calculated that $H^1(\R,U(n)^\ad)$ is in bijection with $\{(p,q)\in\mathbb{N}^2:p+q=n\text{ and }p\geqslant q\}$. By Lemma 2, we know that the groups $U(p,q)$, as $(p,q)$ ranges over such pairs, form non-isomorphic inner forms of $U(n)$. We thus deduce the following.

Corollary 9: The following equality holds $$\mathrm{InnForm}(U(n))=\left\{U(p,q):(p,q)\in \N^2\text{ and }p+q=n\text{ with }p\geqslant q.\right\}.$$

To compute the cohomology of $U(p,q)$ we recall the following definition.

Definition 10: Let $G$ be a reductive group over $\R$. An inner form of $G$ is called pure if its associated class in $H^1(\R,G^\ad)$ (as in Lemma 8) is in $\mathrm{im}\left(H^1(\R,G)\to H^1(\R,G^\ad)\right)$.

We then have the following basic calculation.

Proposition 11 ([Kne, §1.4, Lemma]): Let $H$ be a pure inner form of $G$. Then, there is a natural bijection $H^1(\R,G)\xrightarrow{\sim}H^1(\R,H)$.

Now, by Proposition 7 we have that $H^1(\R,U(n))\to H^1(\R,U(n)^\ad)$ is surjective, and so every inner form of $U(n)$ is pure. So, we deduce that for any unitary group $U$ we have that

$$H^1(\R,U)\simeq \{(p,q)\in \N^2:p+q=n\text{ and }p\geqslant q\},$$

and so, in particular, for each $U=U(p,q)$.

To calculate the inner forms of $\mathrm{GL}_{n,\R}$, we use the exact sequence

$$1\to \mathbb{G}_{m,\mathbb{R}}\to\GL_{n,\R}\to\mathrm{PGL}_{n,\R}\to 1,$$

which, with Hilbert'9 Theorem 90 and Lemma 4, gives us an exact sequence of pointed sets

$$1\to H^1(\R,\mathrm{PGL}_{n,\R})\to \Z/2.$$

It's not hard to show that this map is actually bijective if $n$ is even, and trivial if $n$ is odd (e.g., see [Sza, §4.4]). So, we get that $\mathrm{GL}_{n,\R}$ has no non-trivial inner forms if $n$ is odd, and one if $n$ is even. I leave it for you to check that if $n=2m$, and $A_m:=\mathrm{Mat}_{m\times m}(\mathbb{H})$ (where $\mathbb{H}$ are Hamilton's quaternions) then the units $A_m^\times$ are an inner form of $\GL_{n,\R}$. The Galois cohomology $H^1(\R,A_m^\times)$ is trivial as follows from Generalized Hilbert's Theorem 90 (see [Kne, §1.7, Example 1]).

Thus, all in all, we deduce the following omnibus result.

Proposition 12: There is a decomposition $$\mathrm{Form}(\mathrm{GL}_{n,\R})=\mathrm{InnForm}(\mathrm{GL}_{n,\R})\sqcup \mathrm{InnForm}(U(n)).$$ There are further equalities $$\mathrm{InnForm}(\GL_{n,\R})=\begin{cases} \{\GL_{n,\R}\} & \mbox{if}\quad n\text{ is odd}\\ \{\GL_{n,\R},A_m^\times\} & \mbox{if}\quad n=2m.\end{cases}$$ and $$\mathrm{InnForm}(U(n))=\{U(s,t): (s,t)\in\N^2\text{ and }s+t=n\text{ with }s\geqslant t\}.$$ Furtheremore, we have that $$H^1(\R,\GL_{n,\R})=H^1(\R,A_m^\times),$$ are trivial and $$H^1(\R,U(s,t))\simeq \{(p,q)\in \N^2: p+q=n\}.$$

References

[Bor] Borovoi, M., 2023. Galois cohomology of reductive algebraic groups over the field of real numbers. Communications in Mathematics, 30.

[Kne] Kneser, M. and Jothilingam, P., 1969. Lectures on Galois cohomology of classical groups (No. 47). Bombay: Tata Institute of Fundamental Research.

[Mil] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Sza] Gille, P. and Szamuely, T., 2017. Central simple algebras and Galois cohomology (Vol. 165). Cambridge University Press.