Let $K$ be a CM-field with totally real subfield $F$. Let $(V\_1, h\_1)$ and $(V\_2, h\_2)$ be two $n$-dimensional $K$-vector spaces with nondegenerate Hermitian forms, where $n\geq 3$.



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> **Question 1** Does every $F$-group isomorphism $PU(V\_1, h\_1)\cong PU(V\_2, h\_2)$ arise from an isometry of $(V\_1, h\_1)\cong (V\_2, \lambda h\_2)$ for some $\lambda\in F^{\times}$ ?
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> **Edit. Question 2** Suppose there exists an $F$-group isomorphism $PU(V\_1, h\_1)\cong PU(V\_2, h\_2)$, can we deduce $(V\_1, h\_1)\cong (V\_2, \lambda h\_2)$ for some $\lambda\in F^{\times}$, or any other relations between $h\_1$ and $h\_2$?
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A related statement for orthogonal group is given in the comment to the question [$p$-adic orthogonal groups in four variables](https://mathoverflow.net/questions/226214/p-adic-orthogonal-groups-in-four-variables) .


We are interested in the case that $K$ is a cyclotomic field and $h$ has signature $(1,n-1)$ for an embedding $F\to \mathbb{R}$ and is definite for all the other embeddings.



Does the F-unitary group isomorphism arises from a conformal isometry?
The answer to Question 1 is **No.** Indeed, take $V\_1=V\_2=V:=K^n$ and write
$$ h\_1(z)=h\_2(z)=h(z):=\lambda\_1 z\_1 \bar z\_1+\dots+\lambda\_n z\_n\bar z\_n\quad\text{with}\
\lambda\_i\in F^\times.$$
Write $\widetilde G=U(V,h),\ G={\rm PU}(V,h)$.
Define
$$\tilde\sigma\colon\,\widetilde G\to \widetilde G,\quad g\mapsto \bar g\quad\text{for}\ g\in \widetilde G(F)\subset {\rm GL}(n,K).$$


Then the $F$-automorphism $\tilde \sigma$ of $\widetilde G$
induces an $F$-automorphism $\sigma$ of $G$.
Since $n\ge 3$, this autmorphism $\sigma$ is *outer*, and hence it is not a conformal isometry. The difference with the case of [Brian Conrad's comment](https://mathoverflow.net/questions/226214/p-adic-orthogonal-groups-in-four-variables#comment558491_226214) is that in his case the algebraic group has no outer automorphisms.


**Edit.** The answer to Question 2 is **Yes**.


The isomorphism classes of twisted forms $(V',\,F^\times\cdot h')$ of the Hermitian space $(V,\,F^\times\cdot h)$ bijectively correspond to $H^1(F, {\rm GU}(V,h))$,
where ${\rm GU}(V,h)={\rm Aut}(V,\,F^\times\cdot h)$.
The isomorphism classes of twisted forms of the algebraic $F$-group ${\rm PU}(V,h)$ bijectively correspond to $H^1(F,{\rm Aut}({\rm PU}(V,h)))$.
We wish to show that the kernel
$$\ker\big[ H^1(F, {\rm GU}(V,h))\longrightarrow H^1(F,{\rm Aut}({\rm PU}(V,h)))\big]$$
is trivial.


We factor the above arrow as
$$H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$
(where $^0$ denotes the identity component),
and show that both arrows have trivial kernels.
For the first arrow we have an exact sequence
$$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0),$$
which shows that the kernel is trivial. For the second arrow we have a short exact sequence of $F$-groups,
$$1\to {\rm Aut}({\rm PU}(V,h)^0)\to {\rm Aut}({\rm PU}(V,h))\to {\rm Aut}(A\_{n-1})\to 1,$$
where $A\_{n-1}$ is the corresponding Dynkin diagram.
We obtain a cohomology exact sequence
\begin{align\*}
{\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A\_{n-1})(F)\to H^1(F,&{\rm Aut}({\rm PU}(V,h))^0)\\
&\to H^1(F,{\rm Aut}({\rm PU}(V,h)))
\end{align\*}
where the arrow ${\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A\_{n-1})(F)$
is surjective by the answer to Question 1.
This shows that the arrow
$$H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$
has trivial kernel, which completes the proof.


