Example
What are the equivariant Chern classes of line bundles over a point \(\text{pt}\) with a \(\mathbb C^*\) action?
- A \(\mathbb C^*\)-equivariant vector bundle over \(\text{pt}\) is simply a vector space \(V\) with a \(\mathbb C^*\) action; i.e. a \(\mathbb C^*\)-representation \(\mathbb C^* \to \Aut(V)\).
- If we want a line bundle, then \(\Aut(V) = \mathbb C^*\)and so we’re asking for a character \(a\in \Hom(\mathbb C^*, \mathbb C^*) = \mathbb Z.\) Thus up to isomorphism the equivariant line bundles over a point are the one-dimensional representations \(\mathbb C_a\) of \(\mathbb C^*\), where \(z\cdot v = z^a v\).
- The equivariant Chern class \(c_1^G(E)\) of an equivariant vector bundle \(E\to B\) is the ordinary Chern class \(c_1(\mathbb E\times^{G}E)\) of \(\mathbb E\times^GE\) over \(\mathbb E \times^G B\), so in our case the equivariant Chern class \(c_1^{\mathbb C^*}(\mathbb C_a) := c_1(\mathbb E\times^{\mathbb C^*} \mathbb C_a)\) is an ordinary Chern class of a line bundle over \(\mathbb B = \mathbb E\times^{\mathbb C^*} \text{pt}\).
- In our case, \(\mathbb C^m\setminus \{0\}\) serves as an approximation space \(\mathbb E\). This means \(\mathbb B = (\mathbb C^m\setminus \{0\})\times^{\mathbb C^*} \text{\{pt\}} = \mathbb P^{m-1}.\)
- Since \(\mathbb P^{m-1} = \Gr(1, \mathbb C_1)\), by Exercise 2.1.1. we get an identification between the tautological line bundle \(\mathcal O(-1)\) of \(\mathbb P^{m-1}\) and \((\mathbb C^{m}\setminus \{0\})\times^{\mathbb C^*} \mathbb C_1\). Note that this is only true when we use the standard representation \(\mathbb C_1\), as this matches the left action of \(\GL_1 = \mathbb C^*\) on \(\mathbb C\) in the exercise.
- We therefore get that \[c_1^{\mathbb C^*}(\mathbb C_1) = c_1(\mathcal O_{\mathbb P^{m-1}}(-1)).\] It is a fact from intersection theory that this Chern class generates the cohomology of \(\mathbb P^{m-1}\), so by taking larger and larger approximation spaces we see that for \(\hbar = c_1(\mathcal O_{\mathbb P^{m-1}}(-1))\), \[\Lambda_{\mathbb C^*} = \mathbb Z[t].\]
- Using the fact that \(c_1^G(L\otimes M) = c_1^G(L)+c_1^G(M)\) for \(G\)-equivariant line bundles, we get the identification \[c_1^{\mathbb C^*}(\mathbb C_a) = a\hbar = c_1(\mathcal O(-a)).\] You could alternatively get this from repeating the argument in Exercise 2.1.1 with a modified action on the \(\mathbb C^d\) component.