4
$\begingroup$

Let $C^p(M)$ denote the group of closed $p$-forms on the manifold $M$, and $Z^p(M)$ the group of all exact $p$-forms on the manifold $M$. The de Rham cohomology is given by the quotient,

$$H^p(M)=C^p(M) \, / \, Z^p(M)$$

I am interested in computing the de Rham cohomology of a Schwarzschild manifold, i.e.

$$H^p\left(\mathbb{S}^2 \times \mathbb{R}^2\right)$$

Proceeding from the above definition may be arduous, so could one use a Mayer-Vietoris approach instead? If so, what is the Mayer-Vietoris sequence for cohomology (rather than homology), and what decomposition of the manifold, if any, can we apply to use Mayer-Vietoris? Otherwise I was hoping to utilize a theorem which expresses the cohomology of a Cartesian product of spaces, as $$H^p(\mathbb{S}^2)\quad H^p(\mathbb{R}^2)$$

are well-known cohomologies. Other creative/different approaches to computing the cohomology are certainly welcome and encouraged.

$\endgroup$
5
$\begingroup$

There is such a theorem about product spaces, it's called the Künneth formula. For de Rham cohomology, $$H^n(Y \times X) \cong \bigoplus_{i+j=n} H^j(X)\otimes H^i(Y).$$ Since the cohomology of $S^n$ and $\mathbb R^n$ are both well known, the calculation is now simple. You can find this theorem in the book by Bott and Tu, Differential Forms in Algebraic Topology.

You could of course also use a Mayer-Vietoris sequence with $U$ as a spherical cap slightly south of the equator, and $V$ as a spherical cap slightly north of the equator, both extended for all times. Then both $U$ and $V$ have the topology of $\mathbb R^4$ (because a hemisphere has the topology of $\mathbb R^2$) and $U \cap V$ has the topology of $S^1 \times \mathbb R^3$ (because a band around sphere has the topology of $S^1 \times \mathbb R$). Then you need to find the cohomology of $S^1 \times \mathbb R^3$. You can do this also with the Mayer-Vietoris sequence. Bott and Tu use the M-V sequence on $S^1$ as an example, this calculation will be essentially the same.

There's a good reason that the calculation will be essentially the same, namely that $S^1$ is a deformation retract of $S^1 \times \mathbb R^3$. Cohomology is invariant under deformation retracts, so we really just need to know the cohomology of $S^1$. In fact for the original problem of $S^2 \times \mathbb R^2$ has $S^2$ has a deformation retract, so we could have used this from the beginning.

However calculating the cohomology of $S^n$ is rather simple with the M-V sequence. You just need to know it for $S^1$ and then you can proceed by induction.

$\endgroup$
  • $\begingroup$ Thank you, this is just what I needed! I feel embarrassed now that the problem seems trivial, but I'm just learning about cohomology :) $\endgroup$ – JamalS May 7 '14 at 17:03
  • $\begingroup$ I feel embarrassed that I missed an even simpler argument, please see the updated answer. :) $\endgroup$ – Robin Ekman May 7 '14 at 17:05
  • $\begingroup$ I'm just wondering in GR, for many manifolds which are solutions to a particular Einstein field equation, all we know about them is their metric, Killing vectors, etc. How can we proceed from that point to a point where we can compute global properties, e.g. cohomology? $\endgroup$ – JamalS May 7 '14 at 17:09
  • $\begingroup$ That's a question I've been wondering about myself, you should post it as a question! Maybe some places to start looking could be Exact Solutions by Stephani et al. and The Large-Scale Structure of Space-time by Hawking and Ellis. $\endgroup$ – Robin Ekman May 7 '14 at 17:14
  • $\begingroup$ If you can prove the solution is compact, you could get to the Euler characteristic fairly quickly using the Gauss-Bonnet-Chern theorem. I'll post the question :) $\endgroup$ – JamalS May 7 '14 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.