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In this book (and, in what seems to be an equivalent fashion, in this article), Gel'fand and Fuks calculate the Lie algebra cohomology of the formal vector fields $W_n$ on $\mathbb{R}^n$ with trivial coefficient space $\mathbb{R}$. For those unfamiliar, the general idea can be summarized as follows (I think): They find $\mathfrak{gl}(n,\mathbb{R})$ as a Lie subalgebra of $W_n$ and investigate the arising Hochschild Serre spectral sequence, with first page $$ E^{p,q}_1 = H^q(\mathfrak{gl}(n,\mathbb{K}) ; \Lambda^p(W_n/\mathfrak{gl}(n,\mathbb{K}))^\star).$$ They derive that the first and second page are identical, as the differential on the first page vanishes. They relate the second page of this spectral sequence to the second page of another spectral sequence, namely the Leray-Serre spectral sequence of a certain bundle $$ X_n \to \text{sk}_{2n}\mathbb{C}G(N,n),$$ where:

  • $\mathbb{C}G(N,n)$ is the Grassmanian of complex $n$-planes in $\mathbb{C}^N$, where we take an arbitrary $N \geq 2n$, as the following construction is then independent of $N$.
  • $\text{sk}_{2n}\mathbb{C}G(N,n)$ denotes the $2n$-skeleton of this Grassmanian considered as a CW-complex (using the Schubert cell decomposition)
  • the bundle structure comes from considering the canonical $U(n)$-bundle over the Grassmanian $$\mathbb{C}V(N,n) \stackrel{\pi}{\to} \mathbb{C}G(N,n),$$ where $\mathbb{C}V(N,n)$ denotes the Stiefel manifold of orthonormal $n$-frames in $\mathbb{C}^N$, and as such
  • the total space $X_n$ is defined as $$ X_n := \pi^{-1}(\text{sk}_{2n}\mathbb{C}G(N,n)).$$

That was quite a mouthful. Now, they find that the $E^{p,0}_2$ term of the first spectral sequence is a polynomial algebra mod degree $2n$ where the generators are images under transgression of generators of the exterior algebra $E^{0,q}_2$. From this, they conclude the following, which I cite from the paper (with a few notational adjustments):

The second term of the real cohomology spectral sequence of the bundle $(X_n,\pi,(\mathbb{C}G(N,n))_{2n})$ described in the Introduction is $H^\bullet((\mathbb{C}G(N,n)_{2n};\mathbb{R}) \otimes H^\bullet(U(n);\mathbb{R})$. From Theorem 19.6 of (this French paper) it follows that this tensor product is isomorphic as a bigraded algebra to the term $E_2$ of our spectral sequence, and also that in the two spectral sequences the actions of the differentials are the same. This means that the limits of the spectral sequences are the same, and thus $H^q(W_n;\mathbb{R}) = H^q(X_n,\mathbb{R})$ for all $q$.

I can give further details if it seems necessary, but for now, I think I will leave it at this. My question now is, how do they manage to conclude that the cohomologies are equal? It is true that by this argument both spectral sequences converge to the same $E^\infty$ term, but can one really conclude equality of the cohomologies? Looking at this MO question and this one, this does not seem to be the case, unless the isomorphism had already been induced by a map on the level of cochains, as the isomorphism strictly takes place on $E_2$. Alternatively, it would suffice if the spectral sequence degenerated on some page, but the only further result they prove is $E^{p,q}_\infty = 0$ for $0 < p + q \leq n$ and that multiplication in the cohomology ring is trivial as a consequence, which is not enough for degeneracy by my understanding.

If there is a subtle or non-so-subtle point that I am missing, I would be very thankful for any help. If there is a point that I should elaborate on in more detail, I will gladly do so.

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  • $\begingroup$ Could it just be because $E_1=E_2$? $\endgroup$
    – Jeff Strom
    Commented Aug 3, 2018 at 11:10
  • $\begingroup$ How do you mean? Would there be a way to derive from this fact that there's an isomorphism-inducing map on the chain-level or so? I must admit, I don't have a large body of experience with spectral sequence shenanigans. $\endgroup$
    – user126256
    Commented Aug 3, 2018 at 11:21
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    $\begingroup$ The $E^\infty$ page is the associated graded of cohomology (there don't seem to be any convergence issues). Then if you're looking to calculate rank of cohomology in each degree you are finished, because when you work over a field there are no extension problems; a filtered vector space is determined up to isomorphism by its associated graded. So you just need to read rank off of the diagonals. The ring structure is not completely determined but it seems this is taken care of by another step. $\endgroup$
    – mme
    Commented Aug 3, 2018 at 14:09
  • $\begingroup$ Oh, that's way more simple than I thought! I wasn't aware the extension problem didn't occur for field coefficients. Should've done the calculation myself. Thank you, that was all I needed! $\endgroup$
    – user126256
    Commented Aug 3, 2018 at 14:27
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    $\begingroup$ Ideally, one wants a map $F$ of filtered complexes from the complex that computes $H^*(W_n,\mathbb R)$ to the one that computes $H^*(X_n,\mathbb R)$. In the situation of the authors, where the $E_2$-pages are isomorphic (by this I mean the map induced by $F$ on that page is an isomorphism), you get that the map $F_\infty$ on the $E_\infty$ page is an isomorphism. Since this is just the map $\mathsf{gr}\,H^*(F)$, weak conditions on the filtrations will guarantee that $F$ is a quasi-isomorphism. Else, you can only say dimensions match. The first scenario is of course much more desirable. $\endgroup$
    – Pedro
    Commented Aug 3, 2018 at 16:31

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