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So I'm attempting to compute the $\chi_{y}$ genus of a one-dimensional torus. Even though I know it's certainly not zero, there are two reasons why I convinced myself it was; one of them probably very basic, the other to do with Atiyah-Bott localization.

Before getting there, let $X$ be a compact, complex manifold with formal Chern roots $x_{i}$. The Hirzebruch $\chi_{y}$ genus is given by

$$\chi_{y}(X) = \int_{X} \prod_{i} (1+ye^{-x_{i}}) \frac{x_{i}}{1-e^{-x_{i}}}$$

Now, for a one-dimensional torus, there is only one factor in that product. Moreover, its Chern root is zero since the tangent bundle is trivial. However, one cannot naively set $x=0$, since you end up with a indeterminate $0/0$ expression. Rather, I expanded the integrand:

$$(1+ye^{-x})\frac{x}{1-e^{-x}}=(y+1) + \frac{1}{2} x (y-1) + \cdots$$

Since the integral is a pairing of homology and cohomology classes, we only get contributions from one of those terms leaving,

$$\chi_{y}(X) = \frac{1}{2}(y-1)\int_{X} x $$

If that integral were one, I think that would be the right answer, but why isn't that integral zero!? That Chern root should be zero, and moreover, the integral of the top Chern class is the Euler characteristic which is zero. Where exactly am I overlooking something here?

Localization Argument: The other reason I convinced myself this should vanish, is that a torus acts freely on itself. So I thought that by the Atiyah-Bott localization, if there are no fixed points, then the integral simply vanishes. Am I overlooking something in equivariant cohomology here?

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    You say you have two arguments that ir's zero, but also that you know it's not zero. What's your argument for that? – 2017-02-21
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    Are you reading Hirzebruch's book ? – 2017-02-21
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    Oh geez, as I was responding to this I realized I was doing something dumb. I thought it wasn't zero because I was applying the actual definition of the genus as the alternating sums over Betti numbers. I was adding these Betti numbers up incorrectly, so I was getting something non-zero. Now, it seems like the zero arises there as it should, and it will also vanish for a complex torus of any dimension. Does this sounds right? I was actually using Huybrecht's complex manifolds book, Rene. – 2017-02-21
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    Yes, it vanishes for a complex torus of any dimension, and IMO the easiest way to see this is to observe that the Chern classes, hence the Chern numbers, all vanish. The whole business with Chern roots is just a convenient way to write down a bunch of Chern numbers; that is to say, the Hirzebruch $\chi_y$ genus is a polynomial in $y$ whose coefficients are linear combinations of Chern numbers. – 2017-02-22
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    @QiaochuYuan Makes total sense, thanks a lot! Do you think my localization argument is also valid? Because in that case, something of higher technology like an 'Elliptic Genus' which is also an integral over the torus would vanish. In fact I think it probably does, since this elliptic genus is also simply some crazy linear combination of Chern numbers. – 2017-02-22
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    Yes, the localization argument also works. – 2017-02-22

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