We can define the [signature of a manifold](https://en.wikipedia.org/wiki/Signature_(topology)) 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](https://mathoverflow.net/a/142119/27004).



Signature of the manifold of the multiple fibrations over spheres
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.


