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This is about the complex version of this question,

I am interested in understanding the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^n)$ as the following quotient of the integral cohomology ring of $\mathrm{Gr}_k(\mathbb{C}^{\infty})$: $$ H^*(\mathrm{Gr}_k(\mathbb{C}^n))\cong H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle. $$ Here the $\bar{c}_i$ are polynomials in the Chern classes $c_i$ of the tautological $k$-plane bundle over $\mathrm{Gr}_k(\mathbb{C}^{\infty})$, and are defined by the relation $$ c\bar{c}=1. $$ (Recall that $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\cong \mathbb{Z}[c_1,\dots,c_k].$)

The approach suggested for the real case (with $\mathbb{Z}/2$ coefficients) in the linked question is to show that

  1. the inclusion $\mathrm{Gr}_k(\mathbb{C}^{n})\subset \mathrm{Gr}_k(\mathbb{C}^{\infty})$ induces a surjection $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))\to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;
  2. the polynomials $\bar{c}_{n-k+1},\dots,\bar{c}_n$ are zero in $H^*(\mathrm{Gr}_k(\mathbb{C}^n))$, hence the above map induces a surjection from the quotient $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle \to H^*(\mathrm{Gr}_k(\mathbb{C}^n))$;
  3. use a dimension count to deduce that the latter map is also injective, hence an isomorphism.

Question: Since we are working with integral cohomology, which is merely a ring (not a $\mathbb{Z}/2$-vector space), my question is how one could carry out or alter step (3) to obtain injectivity.

Thanks for the help. Any references are appreciated.

Eric Wofsey
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cork_twist
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1 Answers1

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For each $i$, $H^i(\mathrm{Gr}_k(\mathbb{C}^n))$ is a finitely generated free abelian group, as is the $i$th degree of $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle$. So, if you can show they have the same rank, any surjective homorphism between them is automatically injective as well (any surjection $\mathbb{Z}^n\to\mathbb{Z}^n$ is injective; more generally, this is true for any Noetherian module over a ring, by considering the kernels when you iterate the homomorphism). If you like, this means you can count their dimensions after tensoring with $\mathbb{Q}$ (or equivalently, use cohomology with coefficients in $\mathbb{Q}$) and check that they are the same.

Eric Wofsey
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  • How do you know that $H^i(\mathrm{Gr}k(\mathbb{C}^n))$ and the $i$-the degree of $H^*(\mathrm{Gr}_k(\mathbb{C}^{\infty}))/\langle \bar{c}{n-k+1},\dots,\bar{c}_n\rangle$ are free, i.e. have no torsion? Are you using UCT? – cork_twist Feb 09 '21 at 20:00
  • For $H^(\mathrm{Gr}_k(\mathbb{C}^{n}))$ you can use the Schubert cell structure, which is a CW complex structure with cells only in even degrees (this is also how you can identify the rank). For $H^(\mathrm{Gr}k(\mathbb{C}^{\infty}))/\langle \bar{c}{n-k+1},\dots,\bar{c}_n\rangle$ I imagine you'd have to actually do some algebraic calculation with $\bar{c}$ (you would need to do such a calculation to determine its rank anyways). – Eric Wofsey Feb 09 '21 at 20:06
  • Or, if you already know the homology groups by some other means, sure, you could use the UCT. – Eric Wofsey Feb 09 '21 at 20:07
  • What kind of algebraic calculation do you envisage to determine the rank of $H^*(\mathrm{Gr}k(\mathbb{C}^{\infty}))/\langle \bar{c}{n-k+1},\dots,\bar{c}_n\rangle$? I am able to see from the Schubert cell structure that $H^r(\mathrm{Gr}_k(\mathbb{C}^n))$ is free, and nontrivial only for $r$ even, in which case the rank equals the number of partitions of $r$ into at most $k$ integers, each of which is at most $n-k$ (also proved in Milnor-Stasheff). Is there some trick to determine the Poincare polynomial? – cork_twist Feb 13 '21 at 01:53
  • I don't know off the top of my head. – Eric Wofsey Feb 13 '21 at 01:54
  • I have found a computation of the Poincare polynomial of $\mathbb{Z}[c_1,\dots,c_k]/\langle \bar{c}_{n-k+1},\dots,\bar{c}_n\rangle$ in Bott-Tu's book (it turns out to be a Gaussian polynomial), which allows one to determine the ranks of the degree-$i$ components. However, I am still unsure how to show that the $i$th degree is free. How could one prove this without resorting to Schubert calculus? Thanks for your help. – cork_twist Feb 13 '21 at 20:31
  • If the calculation also works to give the same rank with coefficients in $\mathbb{F}_p$ for each $p$, then that would imply there can't be torsion. (Alternatively, I would not be surprised if you can directly chase through the calculation with coefficients in $\mathbb{Z}$ to find that it is free.) – Eric Wofsey Feb 13 '21 at 20:41