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This is a question about the proof in the book Milnor&Stacheff, computing the cohomology of infinite complex Grassmann $G_n(\Bbb C^\infty)$ by induction on $n$.

Assume $n>1$. Let $E$ denote the total space of the canonical bundle over $G_n(\Bbb C^\infty)$, so that $E=\{(X,v)\in G_n(\Bbb C^\infty)\times \Bbb C^\infty:v\in X\}$, and let $E_0$ denote the subset of $E$ consisting of all pairs $(X,v)$ with nonzero $v$. Then there is a well-defined map $f:E_0\to G_{n-1}(\Bbb C)$ sending $(X,v)$ to the orthogonal complement of $v$ in $X$. The book then asserts that $f$ induces isomorphisms on cohomology ($\Bbb Z$ coefficients), and its proof goes as follows: For sufficiently large $N$, let $f_N$ denote the restriction of $f$ to $E_0(\Bbb C^N):=\{(X,v)\in G_n(\Bbb C^N)\times \Bbb C^N:v\in X, v\neq 0\}$ (thus $f_N$ maps into $G_{n-1}(\Bbb C^N)$). Then show that $f_N$ induces an isomorphism $H^i(G_{n-1}(\Bbb C^N))\to H^i(E_0(\Bbb C^N))$ for $i\leq 2(N-n)$, and "taking the direct limit $N\to \infty$, $f$ induces isomorphisms in all dimensions".

My question is: what does "taking the direct limit" means? I understood the proof that $f_N$ induces isomorphisms on dimension $\leq 2(N-n)$, but I cannot see why $f$ induces isomorphisms on all dimensions.

(Another one: Is the continuity of $f$ obvious? I can't see this neither)

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  • $\begingroup$ Do you know what the direct limit of a sequence of spaces is ? $\endgroup$ Aug 17, 2020 at 8:11
  • $\begingroup$ @MaximeRamzi Do you mean: $A$ has the direct limit topology of $A_1\subset A_2\subset \cdots $ if a set $B\subset A$ is open in $A$ iff $B\cap A_n$ is open in $A_n $ for all $n$? $\endgroup$
    – user302934
    Aug 17, 2020 at 8:17
  • $\begingroup$ Yes. If you know that, I can answer your question $\endgroup$ Aug 17, 2020 at 8:20
  • $\begingroup$ @MaximeRamzi Yes I know that and that $G_n(\Bbb C^\infty)$ and $\Bbb C^\infty $ have the direct limit topologies of $G_n(\Bbb C^n)\subset G_n(\Bbb C^{n+1})\subset \cdots$ and $\Bbb C^1\subset \Bbb C^2\subset \cdots$, respectively $\endgroup$
    – user302934
    Aug 17, 2020 at 8:27

1 Answer 1

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In the comments, you seem to know what the direct topology is.

Here, not that $G_n(\mathbb C^\infty)$ is the direct limit of the $G_n(\mathbb C^N)$.

Moreover, there's an important claim:

Let $X_1\subset ...\subset X_n \subset ...$ be nice inclusions of spaces (you need the inclusions to be closed and the spaces to be $T_1$, assumptions which are obviously satisfied here), and $X$ the direct limit of the $X_i$'s. Then for any $k$, $H_k(X)$ is the direct limit of the $H_k(X_n)$'s.

This is true for maps as well : a map induced on direct limits by a system of maps $f_n : X_n\to Y_n$ induces on homology the direct limit of the maps $f_n$.

An important corollary is as follows:

Suppose $(X_n), (Y_n)$ are two such sequences and you have maps $f_n : X_n\to Y_n$. Suppose $f_n$ induces an isomorphism on homology up to degree $F(n)$ for some function $F$ which tends to infinity with $n$. Then the direct limit map $f:X\to Y$ induces an isomorphism on homology.

Proof: suppose you're interested in $H_k$. Then restrict yourself to $n$'s such that $F(n) \geq k$ (this doesn't change the direct limit), and then $H_k(f)$ is the direct limit of the $H_k(f_n)$'s which are isomorphisms. Hence $H_k(f)$ is an isomorphism.

For cohomology you have to be a bit more careful so here there are two options:

1- The proof that was given by the book at stage $N$ works for homology as well. In this case, well the above argument gives you exactly what you need and you need not worry (of course an isomorphism on homology is an isomorphism on cohomology)

2- The proof that was given works only for cohomology. In this case you have to worry a bit, because the cohomology of the direct limit is not exactly the inverse limit of cohomologies, there's a so-called $\varprojlim^1$-term.

In this case, the easiest solution would be to prove that the inclusion $G_n(\mathbb C^N)\to G_n(\mathbb C^\infty)$ is $k(N)$-connected (for some $k(N)$ that tends to infinity with $N$). This is not true for any sequence $X_1\subset ... \subset X_n\subset ...$ so you would probably have to give a specific argument.

If this is true, then by the Hurewciz theorem and the universal coefficient theorem, the inclusion $G_n(\mathbb C^N)\to G_n(\mathbb C^\infty)$ induces an isomorphism on $H^i$ for $i\leq k(N)$; and you should do something similar with $E_0(\mathbb C^N)\to E_0$. Then that would mean that given $i$, the following diagram commutes :

$\require{AMScd}\begin{CD} H^i(G_n(\mathbb C^\infty)) @>>> H^i(E_0) \\ @VVV @VVV \\ H^i(G_n(\mathbb C^N))@>>> H^i(E_0(\mathbb C^N))\end{CD}$

and for $N$ big enough ($i$ being fixed), both vertical maps are isomorphisms, and the lower horizontal one too; so the upper horizontal one is too.

As for continuity of $f$, by a direct limit argument you can restrict to $\mathbb C^N$. I'll give a very rough sketch of how to get continuity of this map, and let you fill in the details.

Take some $(X,v)\in E(\mathbb C^N)$. Locally around it, $E(\mathbb C^N)$ looks like $U\times \mathbb C^n$ for some neighbourhood $U$ of $X$ in $G_n(\mathbb C^N)$.

You can see that you can choose the identification $V\cong U\times \mathbb C^n$ to respect the hermitian product in fibers, because the Gram-Schmidt renormalization process is continuous.

But then in this neighbourhood, the map looks like $U\times \mathbb C^n \to U\times G_{n-1}(\mathbb C^n)\to G_{n-1}(\mathbb C^N)$, so you just have to prove that these two maps are continuous.

For the first one, it's just $\mathbb C^n\to G_{n-1}(\mathbb C^n), v\mapsto v^\bot$. Locally around $v$, you can choose a continuous orthonormal basis, and then it's easy to see how to prove continuity of this map.

The second one is $(X,Y)\mapsto "Y$ in $X$", given a continuous basis of $X$. You can again reason locally (in $Y$) to get continuity.

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  • $\begingroup$ Thanks. I think I need to use the second option, beacuse the isomorphism $H^i(G_{n-1}(\Bbb C^N))\to H^i(E_0(\Bbb C^N))$ for $i\leq 2(N-n)$ induced by $f_N$ came from an application of Gysin sequence. $\endgroup$
    – user302934
    Aug 17, 2020 at 9:13

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