Strata for the nullcone (from Hesselink's paper)

What i understand about strata for the nullcone is this: (from Mumford's "Geometric Invariant Theory" and Hesselink's paper "Desingularizations of Varieties of Nullforms")

ADDED BY DAVID SPEYER In this setting, we are studying a reductive group $G$ acting on a vector space $V$. We write $T$ for a maximal torus of $G$ and $v$ for a nonzero vector in $V$.

There is a bilinear form on the set $Hom(G_m, T) \otimes \mathbb{R}$ that is invariant under the Weyl group, this bilinear form defines a norm on the set of one-parameter subgroups. Consider the set of one-parameter subgroups $\lambda : \mathbb{R} \rightarrow G$ satisfying the condition $\lambda(t).v = 0$. Also this may mean that there is a morphism $f : A^{1} \rightarrow V$ with $f(0)=0, f(t) = \lambda(t)(v)$ for $t \neq 0$; define $m(v, \lambda)$ to be the multiplicity of the fibre $f^{-1} (0)$. Then the strata are defined via the $\Gamma$ which associates to each $v$ a set of one-parameter subgroups of "shortest length", and also satisfying that $m(v, \lambda) \geq 1$; and two vectors are int the same strata if they have the same set of one-parameter subgroups linked with them.

Questions: When $G = GL_{k}(\mathbb{C})$, how would you define the norm on the space $Hom(G_m, T) \otimes \mathbb{R}$ standard torus $T$ of diagonal matrices? Is there some classification of all one-parameter multiplicative subgroups of $GL_{k}(\mathbb{C})$? And I don't understand how $m(v, \lambda)$ is defined - how can the multiplicity of the fibre $f^{-1}(0)$ be anything other than $1$? Is there some exception in the case where $\lambda(t).v=0$ for all $t$, then what is the multiplicity (is it $0$)? How can the multiplicity be, for instance, $2$?

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I've added a paragraph which I think should clarify the question. vinoth, please feel free to delete or edit if I didn't get this right. – David Speyer Nov 28 '09 at 13:37

Example of how the multiplicity can be greater than $1$: Let $\mathbb{G}_m$ act on $\mathbb{A}^2$ by $t: (x,y) \mapsto (t^2x, t^3 y)$. Let $v$ be any element of $\mathbb{A}^2$ not on the coordinate axes, for example, $(1,1)$. So $\mathbb{G}_m$ maps to $\mathbb{A}^2$ by $t \mapsto (t^2, t^3)$. This extends to a map $\mathbb{A}^1 \to \mathbb{A}^2$, given by the same formula.

I'm not sure what your favorite definition of multiplicity is. Mine is to consider the map of rings in the opposite directions: $k[x,y] \to k[t]$ by $x \to t^2$, $y \to t^3$. The point $(0,0)$ corresponds to the ideal $\langle x,y \rangle$; this maps to the ideal $\langle t^2, t^3 \rangle = \langle t^2 \rangle$ in $k[t]$. Then, by definition, the multiplicity of $f^{-1}(0)$ is the dimension of $k[t]/\langle t^2 \rangle$, which is $2$.

The geometric point here is that the map $\mathbb{A}^1 \to \mathbb{A}^2$ has vanishing derivative at $0$, so the preimage of zero has multiplicity greater than $1$.

You also asked about a Weyl invariant form on $\mathrm{Hom}(\mathbb{G}\_m, T)$, where $T$ is a maximal torus of $GL_n$. Every hom from $\mathbb{G}\_m$ to $T$ is of the form $t \mapsto (t^{w_1}, t^{w_2}, \ldots, t^{w_n})$, where the coordinates on the right hand side are the entries in your diagonal matrix. Let's call this $s(w)$.

The obvious choice of Weyl invariant form is $$\langle s(v), s(w) \rangle = \sum v_i w_i.$$

Technically, I note that $\sum v_i w_i + c (\sum v_i) (\sum w_i)$ would also be Weyl invariant, for any integer $c$. No one ever makes the choice of using nonzero $c$, but I don't know a general principle which would exclude it.

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