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 Post subject: A Projective ModulePosted: Sun Oct 23, 2016 5:42 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
To avoid confusion, let us give the following definitions:

Definitions:
• Let $\Lambda$ be $\mathbb{R}, \mathbb{C}$ or the quaternions $\mathfrak{Q}$ and let $G$ be a topological group. A $\Lambda G$ -space is a finite-dimensional vector space $V$ together with a morphism of topological groups $\theta \ \colon G \longrightarrow Aut(V)$. Such a $V$ is also called a representation of $G$ over $\Lambda$.
• Let $V,W$ be $\Lambda G$ -spaces. A $\Lambda G$ -morphism $f \ \colon V \longrightarrow W$ is a $G$-morphism (i.e. $f$ is compatible with the action of $G$) which is a $\Lambda$-linear map.

Fact: If $\theta \ \colon G \longrightarrow Hom_{\Lambda}(V,V)$ is a representation, then the image of $I = \int_{G} \theta = \int_{g \in G} \theta(g) \in Hom_{\Lambda}(V,V)$is $V^{G}$, the subspace of $G$-invariant elements of $V$.

Assertion: (Using the above fact, show that) If $G$ is a compact topological group and $V$ is a $\Lambda G$-space, then $V$ is a projective module (i.e. for every $\Lambda G$-map $\alpha \ \colon V \longrightarrow Y$ and every surjective $\Lambda G$-map $\beta \ \colon X \longrightarrow Y$ there exists a $\Lambda G$-map $\gamma \ \colon V \longrightarrow X$ such that $\alpha = \beta \circ \gamma$).

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