# Signature of the manifold of the multiple fibrations over spheres

We can define the signature of a manifold in $4k$ dimensions.

1) If I understand correctly, the signature $\sigma$ of the manifold of the product space of spheres would always be zero:

$$\sigma(S^n \times S^m \times S^p \times S^q \times \dots )=0$$

Yes?

2) Am I correct that the signature of the manifold of the (multiple) fibrations over spheres will always be zero?

$$S^n \hookrightarrow X_1 \rightarrow S^m$$

$$X_1 \hookrightarrow X_2 \rightarrow S^p$$

$$X_2 \hookrightarrow X_3 \rightarrow S^o$$

$$\sigma(X_j)=0?$$ (for $j=1,2,3,...$, $m,n,o$ are positive integers.)

If I am wrong, is this statement true in 4-dimensions?

Or, can one give one or two simplest counterexamples in 4-dimensions? (or other dimensions?)

See also this post.

## 2 Answers

The signature of an orientable bundle over a sphere with closed fiber vanishes.

Here is an argument via Novikov additivity. Any bundle over $S^k$ with fiber $N^{4n-k}$ is obtained by gluing two copies of $N \times D^k$ (clutching construction) along $N \times \partial D^k$. Now the signature of $N \times D^k$ is zero, because the homology of the boundary, $N \times \partial D^k$, maps surjectively to the homology of $N \times D^k$.

Novikov additivity says that the signature of the original bundle is the sum of the signatures of the two pieces; hence the signature vanishes (without the need for the general Chern-Hirzebruch-Serre result cited above).

Note that this argument fails if $N$ is not closed, because Novikov additivity requires that you glue along closed components of the boundary. A simple example of this failure would be an oriented $D^2$ bundle over $S^2$. The signature is the sign of the Euler class of the bundle.

The total space of a fiber bundle over a sphere that isn't a circle has zero signature by a result of Chern, Hirzebruch, and Serre that the signature is multiplicative when the fundamental group of the base acts trivially on the cohomology of the fiber.

• Thanks Igor, just in case, is it obvious if the base manifold is a $S^1$ circle, does the result still hold? If not, what will be the alternative statement? – wonderich May 24 '18 at 3:51
• @wonderich: See my answer above; the signature vanishes if the base is a circle. – Danny Ruberman May 24 '18 at 5:12
• "The total space of a fiber bundle over a sphere that isn't a circle" -- you mean the total space isn't a circle or the based manifold space isn't a circle? thanks – wonderich May 24 '18 at 15:14
• I assume that "the base is not a circle". (If the total space is a circle, then the base is a circle or a point). In any case Danny Ruberman gave a better answer which does not need the assumption. – Igor Belegradek May 24 '18 at 15:22