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(This is a generalization of a question I posted a week ago.)

I'm looking at a variety sitting inside the algebraic torus $(\mathbb{C}\setminus 0)^n$ generated by the ideal $I = (*x_1^{\alpha_1} + \cdots + *x_n^{\alpha_1}, \dots, *x_1^{\alpha_k} + \cdots + *x_n^{\alpha_k})$, where the $*$ are generic complex numbers who may be zero. Each of the $k\leq n$ polynomials in the ideal belongs to $\mathbb{C}[x_1,\dots,x_n]$ and is homogeneous with no cross terms.

I want to compute the Euler characteristic (i.e. the alternating sum of Betti numbers) of this variety. My question is the following:

How, if at all, is the Euler characteristic of the variety generated by $I$ related to the Euler characteristic of the linear variety generated by $I' := (*x_1 + \cdots + *x_n, \dots, *x_1 + \cdots + *x_n)$?

My hope is that these two numbers are either equal or differ only by a factor of the $\alpha_i$.

I'm coming to this problem from a combinatorics standpoint, and so I have a hope that this Euler characteristic is actually just counting the number of regions of a certain hyperplane arrangement.

Thanks for anything!

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  • $\begingroup$ Jeff, I don't understand what you mean by "defined over the algebraic torus". Do you mean that the coordinate ring of your variety is ${\bf C}[x_1^{\pm}, \dots, x_n^{\pm}] / I$? $\endgroup$
    – Steven Sam
    Commented Sep 29, 2010 at 0:50
  • $\begingroup$ Hey Steven, I probably phrased something wrong in attempting to speak my nonnative math tongue. What I mean is that the polynomials live formally in $\mathbb{C}[x_1,\dots,x_n]$, but that the solution set of the system of these polynomials is only permitted to come from $(\mathbb{C}\setminus 0)^n$. $\endgroup$
    – Jeffrey Doker
    Commented Sep 29, 2010 at 0:57
  • $\begingroup$ Is there a better way I should phrase "defined over the algebraic torus" above to avoid confusing others? $\endgroup$
    – Jeffrey Doker
    Commented Sep 29, 2010 at 0:59
  • $\begingroup$ Maybe "sitting inside" instead of "defined over"? $\endgroup$
    – Steven Sam
    Commented Sep 29, 2010 at 1:00

1 Answer 1

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Such a variety always has Euler number zero. This is because:

  1. Due to being cut out by homogenous equations, it is invariant under the $\mathbf{C}^*$ action $t.(x_1,\ldots,x_n) \to (tx_1, \ldots, t x_n)$.

  2. This action on $(\mathbf{C} \setminus 0)^n$ has no fixed points.

  3. The Euler number of a variety equipped with a $\mathbf{C}^*$ action is equal to the sum of the Euler numbers of the fixed loci.

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