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Let $G$ be a finite group, $V$ be a real vector space of dimension $N$, and $N > |G|$.

Can you find a representation $\sigma: G \to GL(\text{End}(V))$ such that $\sigma$ is not induced by a representation $\rho: G \to GL(V)$ via $\sigma_g(X)=\rho_g^{-1}X\rho_g$; $g \in G$ and $X \in GL(V)$?

We know each automorphism of $GL(\text{End}(V))$ is inner. That is, for each $g \in G$, there exists $U_g \in GL(V)$ such that $\sigma_g(X)=U_g^{-1}XU_g$. Does this necessitate the existant of $\rho: G \to GL(V)$ such that $\rho_g =U_g$ when $N> |G|$?

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    $\begingroup$ $GL(GL(V))$ does not make sense as far as I know. $\endgroup$ Commented Aug 9 at 20:36
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    $\begingroup$ @CaptainLama I believe it should really be $GL(\operatorname{End}(V))$ $\endgroup$
    – Carmeister
    Commented Aug 9 at 20:41
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    $\begingroup$ Were the answer to the question in your title yes, then there would have to be an infinite sequence of ever-simpler representations. $\endgroup$ Commented Aug 9 at 21:30
  • $\begingroup$ You don't even mean $GL(\text{End}(V))$ but $\text{Aut}(\text{End}(V))$, right? That is I assume you're not interested in representations which don't preserve the multiplication (which clearly cannot be induced by conjugation but for simpler reasons). $\endgroup$ Commented Aug 9 at 23:06
  • $\begingroup$ Would you explain what the differences between GL(algebra) and Aut(algebra) is? I thought they were the same. I once asked which notation is better between $\sigma: G \to GL(S^N)$ and $\sigma: G \to Aut(S^N)$, where $S^N$ is the space of $N \times N$ symmetric matrices, and you said the later is awkward. $\endgroup$
    – khashayar
    Commented Aug 10 at 1:45

1 Answer 1

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I assume you mean $\text{Aut}(\text{End}(V))$. By the Skolem-Noether theorem this is $PGL(V)$, so your question is whether every map $G \to PGL(V)$ lifts to a map $G \to GL(V)$.

This is well-understood. Maps $G \to PGL(V)$ are called projective representations, and whether they lift to $GL(V)$ is controlled by whether a corresponding cohomology class in $H^2(G, K^{\times})$ vanishes. This class is sometimes called the Schur multiplier but that name also refers to several other closely related objects; it should really be called something like the Schur class. It is the cohomology class you get when you try to lift each individual element of $G$ and compare the difference between the lift of $gh$ and the product of the lifts of $g$ and $h$.

The condition on the dimension of $V$ is irrelevant because the Schur class doesn't change when you add a genuine representation.

A typical way to produce nontrivial examples of projective representations is to take genuine representations of a central extension

$$1 \to Z \to H \to G \to 1$$

of $G$ on which the center $Z$ acts by scalars (it's a classical fact that every projective representation of a finite group arises in this way), which means the induced map $H \to PGL(V)$ has kernel $Z$ and hence gives a map $G \to PGL(V)$ which usually won't lift to $GL(V)$.

The smallest nontrivial central extension is the quaternion group $Q_8$, which is a central extension of $V_4$ by $C_2$. It has a unique $4$-dimensional irreducible representation $\mathbb{R}^4 \cong \mathbb{H}$ induced by the inclusion $Q_8 \hookrightarrow \mathbb{H}$, on which the center acts by $-1$. This gives a projective representation $V_4 \to PGL_4(\mathbb{R})$ which does not lift to $GL_4(\mathbb{R})$.

To see this explicitly we can think of $Q_8$ as $\{ \pm 1, \pm i, \pm j, \pm k \}$ as a subgroup of $\mathbb{H}$, with center $\pm 1$, so the quotient by the center is $V_4 = \{ 1, i, j, k \}$ with $i, j, k$ all of order $2$. Any lift of the corresponding projective representation $V_4 \to PGL_4(\mathbb{R})$ to $GL_4(\mathbb{R})$ must lift $i, j, k$ to some nonzero real multiples $pi, rj, qk$, but none of these commute.

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  • $\begingroup$ Can you also provide a counterexample when we work over reals? $\endgroup$
    – khashayar
    Commented Aug 9 at 20:54
  • $\begingroup$ @khashayar: $Q_8$ is also a counterexample over the reals, I've just edited. $\endgroup$ Commented Aug 9 at 20:56
  • $\begingroup$ I see, so helpful. $\endgroup$
    – khashayar
    Commented Aug 9 at 20:57

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