Total dimension of the cohomology of a homogeneous space (or of a graded Tor)

Question

I want to calculate the cohomology ring with rational coefficients of a homogeneous space, but would be happy enough to know its total dimension. Let $G$ be a compact Lie group, $T$ a maximal torus, and $S < T$ a subtorus. One has, via Cartan, a formula $$H^*(G/S) \cong \mathrm{Tor}_{H^*(BG)}\big(\mathbb Q, H^*(BS)\big),$$ and one should be able to factor the map $$H^*(BG) \to H^*(BT) \to H^*(BS).$$ If $\mathrm{rk}\ T = t$ and $\mathrm{rk}\ S = s$, we can find a basis elements $u_1,\ldots, u_t \in H^2(BT)$ and linearly independent elements $v_1,\ldots,v_s \in H^2(BT)$ such that this composition is represented by $$\mathbb{Q}[u_1,\ldots,u_t]^{W_G} \hookrightarrow \mathbb{Q}[u_1,\ldots,u_t] \twoheadrightarrow \mathbb{Q}[v_1,\ldots,v_s].$$ So the nature of the inclusion $S \hookrightarrow T$ and the action of the Weyl group of $G$ determine this Tor object entirely, and I want to use these things to find the $\mathbb Q$-dimension of the Tor, somewhat generally. How should I go about doing so?

Answer 1

And now, the example. I've never seen this particular one worked out, so why not give it a shot?

Consider the homogeneous space $Sp(2)/S^1$ where the $S^1$ is embedded as $\operatorname{diag}(z^a, z^b)$ with $\operatorname{gcd}(a,b) = 1$. (This is the most general type of circle action, up to conjugacy).

We have $H^\ast(Sp(2), \mathbb{Q}) = \Lambda_{\mathbb{Q}}(x_3, x_7)$ where the subscript indicates the degree. Thus, $H^\ast(BSp(2),\mathbb{Q}) = \mathbb{Q}[\overline{x_3}, \overline{x_7}]$ and in the spectral sequence of the reference bundle $Sp(2)\rightarrow E \rightarrow BSp(2)$, we have $dx_1 = \overline{x_1}$ and $dx_2 = \overline{x_2}$.

If we let $w_1$ and $w_2$ be coordinates on the "usual" maximal torus of $Sp(2)$, the Weyl group then acts by permuting $w_1$ and $w_2$ and also flipping one or both signs. It follows that the Weyl group invariant elements in $H^\ast(BTSp(2);\mathbb{Q}) = \mathbb{Q}[\overline{w_1}, \overline{w_2}]$ are generated the elementary symmetric polynomials $\sigma_1(\overline{w_1}^2, \overline{w_2}^2) =\overline{w_1}^2 + \overline{w_2}^2$ and $\sigma_2(\overline{w_1}^2, \overline{w_2}^2) = \overline{w_1}^2\overline{w_2}^2.$

We may, thus, identify $H^\ast(BSp(2))$ as the subalgebra $\mathbb{Q}[\overline{w_1}^2 + \overline{w_2}^2, \overline{w_1}^2\overline{w_2}^2]$.

Also, note that $BS^1 = \mathbb{C}P^\infty$, so $H^\ast(BS^1;\mathbb{Q}) = \mathbb{Q}[\overline{z}]$ where $z$ is the coordinate on $S^1$ and $\overline{z} = dz$ in the spectral sequence $S^1\rightarrow E\rightarrow BS^1$.

Now, back to the inclusion $i:S^1\rightarrow Sp(2)$. Letting $z$ denote the coordinate on $S^1$, the induced map $i:H_1(S^1) \rightarrow H_1(T_{Sp(2)})$ maps $z$ to $aw_1 + bw_2$. Dualizing this, it follows that $i^\ast:H^1(T_{Sp(2)})\rightarrow H^1(S^1)$ is given by $i^\ast(w_1) = az$ and $i^\ast(w_2) = bz$.

Thus, we have $Bi^\ast:H^\ast(BT_{Sp(2)}) \rightarrow H^\ast(BS^1)$ given by $Bi^\ast(\overline{w_1}) = a\overline{z}$ and $Bi^\ast(\overline{w_2}) = b\overline{z}$. It follows that $$Bi^\ast(\overline{x_3}) = Bi^\ast(\overline{w_1}^2 + \overline{w_2}^2) = (a^2 + b^2)\overline{z}^2$$ and $$Bi^\ast(\overline{x_7}) = Bi^\ast(\overline{w_1}^2\overline{w_2}^2) = a^2b^2 \overline{z}^4.$$

Now, lets look at the spectral sequence for the main bundle $Sp(2)\rightarrow X\rightarrow BS^1$, keeping in mind that $X$ is homotopy equivalent to $Sp(2)/S^1$.

By naturality, we have $dx_3 = Bi^\ast(\overline{x_3}) = (a^2+b^2) \overline{z}^2$ and $dx_7 = a^2b^2 \overline{z^4}$, on appropriate pages, where $dx_3 = dx_7 = 0$ for all other differentials. So, the $E^2$ page equals the $E^3$ page for trivial reasons. Since $\operatorname{gcd}(a,b) = 1$, it follows that $a^2 + b^2 \neq 0$, so $d:E^3_{0,3}\rightarrow E^3_{4,0}$ is injective, and therefore surjective. Further, $d:E^3_{4,3}\rightarrow E^3_{8,0}$ is an isomoprhism.

So, on $E^4$, everything is zeroed out except $E^4_{i,j}$ with $i = 0,2$ and $j = 0,7$. We know that all differentials must vanish until $E^7$, where $dx_7 = a^2b^2\overline{z}^4$. Now, we already know $\overline{z}^4 = 0$, so this is the trival map. In particular, $E^7 = E^\infty$, so we see (via Poincare duality) that $H^\ast(G/H;\mathbb{Q}) = H^\ast(S^2\times S^7)$ as rings. In particular, the dimension of $H^\ast(G/H;\mathbb{Q})$ is $4$.

One final note - almost everything above works just as well integrally as it does rationally. The only change comes in studying the spectral sequence: $dx_3 = (a^2 + b^2) \overline{z}^2$ is still injective, but no longer surjective. It follows that $E^4(8,0) = \mathbb{Z}/(a^2 + b^2)$ is torsion, and so $dx_7$ has a chance of being nonzero, depending on what $a$ and $b$ are. That said, because $\operatorname{gcd}(a^2 + b^2, ab) = 1$ for relatively prime $a$, and $b$, $E^8_{8,0} = E^\infty_{8,0} = 0$ no matter what. If I've done my math property, it seems as though $H^\ast(G/H;\mathbb{Z}) = \mathbb{Z}[\overline{z}, x_7]/I$ where $I$ is the ideal generated by $(a^2 + b^2) \overline{z}^2$, $\overline{z}^4$, $x_7\overline{z}^2$, and $x_7^2$.

Answer 2

There is probably a faster way to achieve this result, but I'm not immediately seeing it. Note that for equal rank $H$ in $G$, one has that the Poincare polynomial of $G/H$, $p_{G/H}(t)$, is given by the quotient $\dfrac{p_{BH}(t)}{p_{BG}(t)}$ where $BH$ and $BG$ are the classifying spaces of $H$ and $G$. One you have this, $p(1)$ computes the total dimension.

For the case you're considering, I don't know of a faster way than the following, which works for any homogeneous space $G/H$ with $G$ and $H$ compact. This post will contain all the theory, and I'll use another post to actually compute an example.

First, let $E$ denote a contractible space on which $G$ (and therefore $H$) acts freely.

Recall the Borel construction: let $X = G\times_{H} E$. Then projection onto the first factor $\pi_1:X\rightarrow G/H$ is a fibration with fiber $E$. Since $E$ is contractible, $\pi_1$ is actually a homotopy equivalence. Thus, to study the topology of $G/H$, we may as well study that of $X$.

Projection onto the second factor $\pi_2:X\rightarrow E/H = BH$ is a principal $G$-bundle (where $G$ acts on $X$ by left multiplying the $G$ factor). We will use this bundle, which I'll call the main bundle, to study the topology of $X$, and therefore of $G/H$.

Consider the reference bundle $G\rightarrow EG\rightarrow BG$. The main point is the inclusion map $i:H\rightarrow G$ induces a map $Bi:BH\rightarrow BG$ for which the main bundle is the pull back of the reference bundle. (The map on total spaces is given by $\phi:X\rightarrow E$ with $\phi([g,e]) = ge$). Thus, if we can a) understand the differentials in the spectral sequence $G\rightarrow E \rightarrow BG$ and b) understand the pull back map $Bi^\ast$ on cohomology, we can understand the spectral sequence for the main bundle.

a) is simple under your current hypothesis. Rationally, every Lie group has the homotopy type of a product of odd spheres. In particular, $H^\ast(G;\mathbb{Q}) = \Lambda_{\mathbb{Q}}[x_1, x_2,..., x_n]$. The degrees of the $x_i$s are known for each of the simple Lie groups, and this allows one to compute it for any compact group. (These degrees don't change when passing to covers). Now, by studying the spectral sequence for the reference bundle, one easily sees that $H^\ast(BG;\mathbb{Q}) = \mathbb{Q}[\overline{x_1},...,\overline{x_n}]$ with $\deg(x_i) + 1 = \deg(\overline{x_i})$. Further, in an appropriate page in the spectral sequence, $dx_i = \overline{x_i}$ with $dx_i = 0$ on all other pages.

b) is also not so bad. Choose maximal tori $T_H$ of $H$ and $T_G$ of $G$ with $i(T_H)\subseteq T_G$. The induced map $i^\ast:H^\ast(T_G)\rightarrow H^\ast(T_H)$ is easy to compute. As in part a), we see $H^\ast(BT_H)$ is a polynomial algebra in dim$(T_H)$ generators of degree $2$, and similarly for $H^\ast(BT_G)$. Also, we may view the generators of $H^\ast(BT_H)$ as the differential of the generators of $H^\ast(T_H)$ in the spectral sequence $T_H\rightarrow E\rightarrow BT_H$.

The upshot is that we may then compute $Bi^\ast:H^\ast(BT_G)\rightarrow H^\ast(BT_H)$ easily. Now, we want to turn this into information about $BH$ and $BG$ (which, based on your question, you already seem to be familiar with).

The point is that the inclusion $T_H\rightarrow H$ induces a map $H^\ast(BH)\rightarrow H^\ast(BT_H)$. With "nice" coefficients (and $\mathbb{Q}$ is "nice" for all Lie groups), this map is injective with image $H^\ast(BT_H)^{W_H}$ given by the Weyl group invariant elements (the Weyl group acts on $T_H$, hence on $H^\ast(T_H)$, and hence, via differentials, on $H^\ast(BT_H)$.)

Thus, to compute $Bi^\ast:H^\ast(BG)\rightarrow H^\ast(BH)$, you compute it on the $BT_G$ level, and then restrict!

All of this allows you to compute all the differentials in the spectral sequence for the main bundle. Thus, you can, modulo extension problems, completely recover the cohomology ring of $G/H$. In particular, you can always recover the dimension.