Let $k$ be a field, and let $k^\times$ denote its multiplication group. Further let $\mathrm{PGL}(V,k)$ denote the projective general linear group of some vector space $V$ over the field $k$. Denote by $\mathrm{diag}:k^\times\to \mathrm{GL}(V,k)$ the map $a\mapsto a\mathrm{id}_V$. Given an arbitrary section $s:\mathrm{PGL}(V,k)\to \mathrm{GL}(V,k)$ of the canonical projection map $P$ with $s(\mathrm{id}_V)=\mathrm{id}_V$, a projective representation $\rho:G\to \mathrm{PGL}(V,k)$ of some group $G$ induces a unique cohomology class via the 2-cocycle: $$\omega(g,h)=s(\rho(g))s(\rho(h))s(\rho(gh))^{-1}.$$ Generally speaking, a 2-cocycle $\omega:G\times G\to k^\times$ defines a central extension $k^\times\times_\omega G$ of $G$, which, as a set, is defined simply as $k^\times\times G$, but is endowed with the operation: $$(a,g)(b,h)=(ab\omega(g,h),gh).$$

I am trying to prove the following result:

Lemma. Let $G$ be a group. The homomorphisms $G\to\mathrm{PGL}(V,k)$ (i.e., projective representations) inducing (up to equivalence) a 2-cocycle $\omega:G\times G\to k^\times$ are in bijective correspondence to linear representations $\sigma:k^\times\times_\omega G\to\mathrm{GL}(V,k)$ of a central extension, with the property that for all $a\in k^\times: \sigma(a,1_G)\in\mathrm{diag}(k^\times)$.

The idea for the proof is as follows.

Construction 1: Given a projective representation $\rho:G\to \mathrm{PGL}(V,k)$, define $u=s\circ \rho:G\to \mathrm{GL}(V,k)$. Now $u$ satisfies $$u(g)u(h)=\omega(g,h)u(gh),$$ where $\omega$ is the cocycle induced by $\rho$, as described above. One easily verifies that the map $$\sigma:k^\times\times_\omega G\to\mathrm{GL}(V,k);\quad (a,g)\mapsto au(g)$$ defines a representation of the central extension $k^\times\times_\omega G$ that satisfies the desired property.

Construction 2: On the other hand, if given a representation $\sigma:k^\times\times_\omega G\to\mathrm{GL}(V,k)$ with the desired property, define $u:G\to\mathrm{GL}(V,k)$ by $u(g)=\sigma(1_k,g)$. Applying the projection map $P$ after $u$ now gives a projective representation $\rho=P\circ u:G\to\mathrm{PGL}(V,k)$. This construction is easily seen to be the left inverse of Construction 1.

My problem is in proving that Construction 1 is the left inverse of Construction 2. For that, let $\sigma:k^\times\times_\omega G\to \mathrm{GL}(V,k)$ be a representation with the desired property, and let $\rho=P\circ u$ be the induced projective representation. We now apply the first construction to $\rho$, which first gives a map $v:G\to\mathrm{GL}(V,k)$ given by $v(g)=s(\rho(g))=(s\circ P)(\sigma(1_k,g))$, and in turn a map $\kappa:k^\times\times G\to\mathrm{GL}(V,k)$ defined by $\kappa(a,g)=av(g)=a(s\circ P)(\sigma(1_k,g))$. I see no reason why $\sigma=\kappa$.

What am I missing? I feel like I'm overlooking something obvious, or maybe the lemma needs to be tweaked? Any hints are appreciated!


1 Answer 1


Here is an idea. I'll write "$\mathrm{diag}(a)$" instead of "$a$" to emphasize that we are working in $\mathrm{GL}(V,k)$. In your argument, showing that $\sigma = \kappa$ amounts to showing that $x_{1} := \mathrm{diag}(a) \cdot (s \circ P)(\sigma(1_{k},g))$ and $x_{2} := \sigma(a,g)$ are equal for all $a \in k^{\times}$ and $g \in G$. It'd be nice if we can show that the two expressions are both equal to the intermediate expression $x_{3} := \mathrm{diag}(a) \cdot \sigma(1_{k},g)$. With this in mind, perhaps we can modify the statement of the Lemma so that we restrict to group homomorphisms $\sigma : k^{\times} \times_{\omega} G \to \mathrm{GL}(V,k)$ satisfying the following two conditions:

  1. For all $a \in k^{\times}$ we have $\sigma(a,1_{G}) = \mathrm{diag}(a)$, not just $\sigma(a,1_{G}) \in \mathrm{diag}(k^{\times})$. Since $(a,1_{G}) \cdot (1_{k},g) = (a,g)$ in $k^{\times} \times_{\omega} G$, this will imply that $x_{2} = x_{3}$.
  2. For all $g \in G$ we have $(s \circ P)(\sigma(1_{k},g)) = \sigma(1_{k},g)$. This will imply that $x_{1} = x_{3}$.

Note that any $\sigma$ obtained as a result of applying Construction 1 to a projective representation $\rho : G \to \mathrm{PGL}(V,k)$ satisfies the above two conditions.

Then I think we can say the following:

Let $k$ be a field, let $V$ be a $k$-vector space, let $\mathrm{diag} : k^{\times} \to \mathrm{GL}(V,k)$ and $P : \mathrm{GL}(V,k) \to \mathrm{PGL}(V,k)$ be the canonical maps. Fix a section $s : \mathrm{PGL}(V,k) \to \mathrm{GL}(V,k)$ of $P$. Let $G$ be a group and fix a $2$-cocycle $\omega : G \times G \to k^{\times}$. Then there is a bijective correspondence between (1) group homomorphisms $\rho : G \to \mathrm{PGL}(V,k)$ such that $s(\rho(g))s(\rho(h)) = \omega(g,h)s(\rho(gh))$ for all $g,h \in G$ and (2) group homomorphisms $\sigma : k^{\times} \times_{\omega} G \to \mathrm{GL}(V,k)$ such that for all $a \in k^{\times}$ we have $\sigma(a,1_{G}) = \mathrm{diag}(a)$ and for all $g \in G$ we have $(s \circ P)(\sigma(1_{k},g)) = \sigma(1_{k},g)$.

We can probably phrase this in a nice way using the exact sequences $$ 1 \to k^{\times} \stackrel{\mathrm{diag}}{\to} \mathrm{GL}(V,k) \stackrel{P}{\to} \mathrm{PGL}(V,k) \to 1 $$ and $$ 1 \to k^{\times} \to k^{\times} \times_{\omega} G \to G \to 1 $$ of groups, since we are trying to describe the preimage of a single element of $\mathrm{H}^{2}(G,k^{\times})$ under the boundary map $\mathrm{H}^{1}(G,\mathrm{PGL}(V,k)) \to \mathrm{H}^{2}(G,k^{\times})$. There is a seven-term exact sequence of pointed sets $$ 1 \to (k^{\times})^{G} \to (\mathrm{GL}(V,k))^{G} \to (\mathrm{PGL}(V,k))^{G} \to \mathrm{H}^{1}(G,k^{\times}) \to \mathrm{H}^{1}(G,\mathrm{GL}(V,k)) \to \mathrm{H}^{1}(G,\mathrm{PGL}(V,k)) \to \mathrm{H}^{2}(G,k^{\times}) $$ for any group $G$; maybe there is an analogue of the Hochschild-Serre spectral sequence for pointed sets?

  • $\begingroup$ Thank you very much! Wouldn't it be possible to change the section $s$ so that the second condition is fulfilled? (Of course, this will make $s$ dependent on $\sigma$.) Since the central extension is uniquely determined (up to equivalence) by $\rho$, we would still be classifying the linear representations of the same central extension. $\endgroup$
    – Nesta
    Jul 6, 2017 at 10:07
  • $\begingroup$ Hi @Nesta, I've edited my answer a little. I don't know what the correct answer should be if we want to allow $s$ to vary. $\endgroup$ Jul 8, 2017 at 19:55
  • $\begingroup$ Thank you for the extra information; I'm not familiar with the Hochschild-Serre spectral sequence, but I will take a look at it. About the sections: since the cohomology class is independent on the section, we should be albe to tweak it (except for the constraint $s(\mathrm{id})=\mathrm{id}$). $\endgroup$
    – Nesta
    Jul 10, 2017 at 19:42

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