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Suppose I have a smooth manifold $X$ and a Lie group $G$ which acts smoothly on $X$. I'm interested in finding vector bundles $\pi:V\to X$ that inherit the action of $G$, in the sense that there is a smooth action of $G$ on $V$ such that:

  1. If $\pi(v)=x$, then $\pi(gv)=gx$. Or, to put it another way, the action of an element $g$ maps $\pi^{-1}(x)$ into $\pi^{-1}(gx)$.
  2. For each $g\in G$ and $x\in X$, the map $v\mapsto gv$ for $v\in\pi^{-1}(x)$ is a linear isomorphism of the vector spaces $\pi^{-1}(x)$ and $\pi^{-1}(gx)$.

This is motivated by the desire to extend the notion of a group representation to a vector bundle. Note that if $X$ is a single point, then $V$ is just a representation of $G$. A couple examples that work for general $X$:

  1. The trivial line bundle $V=X\times\mathbb R$, along with the trivial action $g(x,a)=(gx,a)$.
  2. The tangent bundle, $V=TX$, with the action of an element $g$ on $TX$ given by the differential of the map $x\mapsto gx$.
  3. Given any two vector bundles $V$ and $W$ with such an action, their direct sum $V\oplus W$ (with action $g(v,w)=(gv,gw)$) and tensor product $V\otimes W$ (with action $g(v\otimes w)=gv\otimes gw$).

Is this something that has been studied before, and if so, where can I learn more about it? I've been reading a bit about principle bundles and $G$-structures, but as far as I can tell those aren't quite the same as what I'm describing (although if they are, I'd appreciate an explanation as to how!).

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    $\begingroup$ What you describe seems to be what is known as an equivariant bundle. $\endgroup$ – Tobias Kildetoft Nov 16 '17 at 9:10
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As the comment notes, this is exactly the notion of a $G$-equivariant vector bundle on the $G$-space $X$. These have been studied extensively before, for the following reason: whenever $\pi: E \to X$ is a $G$-equivariant vector bundle, the vector space of sections $\Gamma(\pi)$ is naturally a representation of $G$. This is not hard to see: let $s: X \to E$ be a section, and define the action of $g \in G$ on $s$ to be $(g \cdot s) = g s(g^{-1} x)$. This is easily checked to be a group action, linear by the hypotheses.

One of the key places this construction is used is to define induced representations in the category of algebraic groups or Lie groups. Let $H \hookrightarrow G$ be a closed (algebraic/Lie) subgroup of $G$. $H$ acts on the right of $G$ by $g \cdot h = gh$. The quotient by this action gives the principal $H$-bundle $\sigma: G \to G / H$, which is $G$-equivariant on the left.

Suppose that $V$ is a representation of $H$, then we may form the associated bundle to $\sigma$, written $\pi: G \times_H V \to G/H$. This is straightforward to explicitly describe: the elements of $G \times_H V$ are pairs $[g, v]$ with the equivalence relation $[gh, v] = [g, hv]$ for any $h \in H$, and the map $\pi$ sends $[g, v] \mapsto gH$. The bundle $\pi$ still has a $G$-equivariant action $g_1 \cdot [g_2, v] = [g_1 g_2, v]$, and so its sections are a representation of $G$, which is (for some people) taken as the definition of the induced representation $\mathrm{ind}_H^G V$.

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