# What condition makes unitary reductive group unramified?

I am a little bit confused with the definition of an unramified unitary group.

Let $F$ be a local field of characteristic zero whose residue field is finite field of characteristic $p$.

Then for a connected reductive group $G$ defined over $F$, recall that $G$ is unramified if it is quasi-split and split over maximal unramified extension of $F$.

Let $E/F$ be an unramified quadratic field extension and $V$ is a hermitian space over $E$. Then unitary group $U(V)$ which is a subgroup of $GL_E$ perserving hermitian form of $V$.

Then I am wondering whether $U(V)$ is unramified reductive group. If not, what conditions ensures $U(V)$ unramified?

Any comments will be highly appreciated.

• The unitary group $G=U(V)$ is a connected reductive group over $F$, and it splits over the unramified quadratic extension $E/F$. It follows that $G$ splits over a maximal unramified extension of $F$. Thus, according to your definition, $G$ is unramified if and only if it is quasi-split. Aug 4 '17 at 3:39
• Thank you for your answer. May I ask you a stupid question? Is the condition that a reductive group G is quasi-split over F equivalent to it splits over some algebraic extension of F? Aug 4 '17 at 9:34
• @Monty, no. Every reductive group splits over an algebraic (even separable) extension. Aug 4 '17 at 9:42
• (cont.) By local class field theory (overkill!) there exists $\lambda\in F^\times$ such that there is no $z\in E$ with $N(z)=\lambda$. Then the hermitian form $$h(z_1,z_2)=z_1 z_1^\rho-\lambda z_2 z_2^\rho$$ does not represent 0 nontrivially. Indeed, if $h(z_1,z_2)=0$ and $z_2\neq 0$, then $N(z_1/z_2)=\lambda$, contradiction. It follows that the group $G=U(E^2,h)$ is anisotropic and therefore, $G$ is not quasi-split. Aug 4 '17 at 11:09
• However, if $d$ is odd, $d=2k+1$, then $h$ has index $k$, and I think that $G=U(V,h)$ must be quasi-split. Aug 4 '17 at 11:19

As Mikhail Borovoi explained in a comment, the question reduces to "when is a unitary group over a non-Archimedean local field quasi-split"? The answer does not distinguish between ramified or unramified separable quadratic extensions.

Let $E/F$ be a separable quadratic extension of non-Archimedean local fields. Isomorphism classes of pairs $(V,h)$ where $V$ is a finite-dimensional vector space over $E$ and $h$ is a non-degenerate hermitian form on $V$ are classified by $(\dim V, \mathrm{disc}(h))$ where $\mathrm{disc} (h) = \det H \mod N_{E/F}(E^{\times})$ with $H$ the matrix of $h$ in some basis of $V$ over $E$. This is the analogue of the classification of quadratic forms (see Jacobson, A note on hermitian forms https://projecteuclid.org/euclid.bams/1183502551 ). In positive dimension every discriminant in $F^{\times} / N_{E/F}(E^{\times}) = \mathbb{Z} / 2 \mathbb{Z}$ occurs.

As Mikhail Borovoi pointed out, in odd dimension these two forms $h_1,h_2$ have the same unitary group, in fact $h_2 \simeq \lambda h_1$ where $\lambda \in F^{\times} \smallsetminus N_{E/F}(E^{\times})$.

In even dimension only one of the two forms gives rise to a quasi-split unitary group, i.e. Mikhail Borovoi's counterexample generalizes to arbitrary even dimension (although in even dimension greater than 2 the group is not anisotropic). There are probably several ways to see this. For example you can show that Borel subgroups of $U(V,h)$ defined over $F$ correspond bijectively to flags $V_1 \subset \dots \subset V_n$ where $\dim_E V = 2n$, $\dim_E V_1 = 1$, $\dim_E V_{i+1}/V_i = 1$ for $1 \leq i \leq n-1$, and $V_n$ is totally isotropic. Or you can argue by taking Galois cohomology of the short exact sequence of algebraic groups over $F$ (for the étale topology) $$U(1) \rightarrow U(V,h) \rightarrow PSU(V,h)$$ and using $H^2(F, U(1)) = 0$ (a consequence of Tate-Nakayama) and the fact that the only inner form of a quasi-split group that is also quasi-split is the trivial one, which follows from the existence of a pinning defined over $F$.

In fact all of this can be understood in the context of Galois cohomology of reductive groups. For example the classification of hermitian forms is essentially equivalent to the Hasse principle (vanishing of $H^1$ of a special unitary group).

By your given definition, for the unitary group you describe to be quasisplit, one would need a Borel subgroup to be defined over the ground field. Since this would correspond to a flag of totally isotropic subspaces (subspaces on which the Hermitian form $v \mapsto h(v, v)$ identically vanishes), this need not happen in general, as there is no reason for a general Hermitian form to have any nontrivial zeros. In general, this group need not be unramified.

• I think you mean 'isotropic'. This answer is the same as @MikhailBorovoi's comments, no? Aug 4 '17 at 3:45
• Yep - thanks for spotting the typo! Aug 4 '17 at 4:01
• There is a reason for $h$ to have nontrivial zeros if the dimension of $V$ over $E$ is greater than 2: $F$ is a nonarchimedean local field. Aug 4 '17 at 4:30