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This question is based on my unanswered post here.

Let $K$ be a field equipped with an involution $\bar{}$, which may be the identity. Suppose $(V,b)$ consists of a $K$-vector space $V$ together with a nondegenerate bilinear or sesquilinear form $b: V \times V \to K$, depending on whether $\bar{}$ is nontrivial or trivial. The endomorphism algebra $\operatorname{End}_K(V)$ admits an adjoint involution with respect to $b$ whenever $b$ is symmetric or skew-symmetric. In this setting, we refer to $(V,b)$ as a reflexive space.

A representation $\rho$ of a finite group $G$ on a reflexive space $(V,b)$ is called isometric if it is $b$-invariant, i.e., for all $g\in G$ , $$b(\rho_gu,\rho_gv)=b(u,v)\ \ ; \forall u, v \in V.$$

This representation extends to $*$-algebra representation $$\rho: KG \to \operatorname{End}_K(V)$$ from the group algebra $KG$ with canonical involution to the endomorphism algebra.

If an irreducible-representation decomposition of $V$ is given by $$V \cong \bigoplus_{i=1}^k W_i^{n_i},$$ then the Wedderburn–Artin theorem implies: $$\operatorname{End}_{KG}(V) \cong \bigoplus_{i=1}^k \operatorname{Mat}_{n_i}(\operatorname{End}_{KG}(W_i))\cong \bigoplus_{i=1}^k \operatorname{End}_{KG}(W_i) \otimes \operatorname{Mat}_{n_i}(K).\label{ISO}\tag{ISO}$$

How can I prove \eqref{ISO} is indeed a $*$-algebra isomorphism if the canonical involution $*$ in $KG$ is anisotropic?

Clarification:

Here, $KG$ is a semisimple group algebra, and the involution is defined with respect to $\bar{}$. That is, for every $a \in KG$:

$$a^*=(\sum_ga_gg)^*=\sum_g\overline{a_g}g^{-1}.$$

The involution $\ast$ is called anisotropic if $a^*a=0$ implies $a=0$ for every $a \in KG$. An anisotropic involution ensures that each simple component $\mathcal{A}_i$ of $KG$ is invariant under the involution, making every centrally primitive idempotent $e_i$ symmetric ($e_i^*=e_i$), as $\mathcal{A}_i\mathcal{A}_j=0$ for $i \ne j$. The involution can then be restricted to each simple factor, $\ast\big|_{\mathcal{A}_i}$. Furthermore, an anisotropic involution guarantees the existence of a symmetric primitive idempotent in each simple component, a property that does not generally hold for arbitrary involutions. So, for each simple left $KG$-module $W$, there exists symmetric primitive idempotent $p$ such that $KGp=W$. Since $pKGp \cong \text{End}_{KG}(W)^{op}$ as $K$-algebra, we can restrict $\ast$ on $KG$ to $\text{End}_{KG}(W)$. This gives the involution on the right-hand side of (ISO), twisting transpose $\bar{}$ with involution on $\text{End}_{KG}(W_i)$.

For the left-hand side, the adjoint involution on $\text{End}_{KG}(V)$ is the map $\tau \mapsto \tau^*$ satisfying $$b(u,\tau v)=b(\tau^*u,v)\ \ \text{ for all }u,v \in V.$$

Now, I have two questions:

1- Does there exist an isomorphism in (ISO) that preserves involution?

2- Does every isomorphism in (ISO) preserve involution?

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    $\begingroup$ You haven't really given us enough information here. What is your "adjoint involution" on $\mathop{\rm End}_K(V)$? Are you assuming that the characteristic of $K$ does not divide $|G|$ so that the decomposition of $V$ given later makes sense? How are $KG$ and $\mathop{\rm End}_K(V)$ made into $*$-algebras? In (ISO), what $*$-algebra structure are you putting on each side? What do you mean by $*$ being anisotropic on $KG$? $\endgroup$ Commented Apr 30 at 21:29
  • $\begingroup$ Does the new edit answer your questions? Please let me know if you want more clarification. $\endgroup$ Commented May 1 at 4:09

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You might want to insist that in $K$ you have $\sum_i\overline{x_i}\, x_i=0$ implies that $\forall i$ $x_i=0$, otherwise anisotropic is a bit of a peculiar condition. In particular, this implies that $K$ has characteristic zero, and it implies that $KG$ is always anisotropic with your definition. This is because the coefficient of the identity in $a^*a$ is $\sum_g \overline{a_g}\, a_g$.

The Wedderburn-Artin isomorphism depends on a choice of basis for each isomorphism class of irreducible over its extended skew-field. The isomorphism (ISO) preserves the involution if and only if the choice of basis is orthonormal in each case with respect to the involution on the extended skew-field. In other words, you want the sum of $n_i$ copies of $W_i$ to be an orthogonal decomposition.

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  • $\begingroup$ For example, over $\mathbb{C}$ with trivial involution, according to your definition $\mathbb{C}G$ is anisotropic if and only if $G$ is an elementary abelian $2$-group. $\endgroup$ Commented May 1 at 9:50
  • $\begingroup$ How can we guarantee the existence of such a basis in each case when the $(KG,*)$ is anisotropic? $\endgroup$ Commented May 2 at 23:40
  • $\begingroup$ What you wrote in the first paragraph is equivalent to saying that the fixed field $K_0$ of $\bar{}$ is formally real. Isn't it? So are you saying $(KG,*)$ is anisotropic if and only if $K_0$ is formally real, and so $b$ acts like an inner product? $\endgroup$ Commented May 2 at 23:50
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    $\begingroup$ The answer to your first question is Gram-Schmidt. Yes, the condition in the first paragraph is saying it's formally real. But it is not true that $(KG,*)$ is anisotropic if and only if $K_0$ is formally real, by the example of my first comment. Even if $G$ has elements that don't square to the identity it's not true. For example if $K=\mathbb{Q}(\omega)$ where $\omega$ is a primitive cube root of unity, with trivial conjugation, and $G$ is cyclic of order four, then $KG$ is anisotropic. So it's a delicate condition, but for not terribly interesting reasons somehow, so use my first paragraph. $\endgroup$ Commented May 3 at 8:51
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    $\begingroup$ I know that free advice never comes cheap, but my advice is to stick with the best generality rather than trying to be as general as you can possibly be, at the sacrifice of a great deal of time and effort, without its being of any use to anyone because there are no good examples. $\endgroup$ Commented May 4 at 18:49

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