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What does the symbol $\#^n S^2 \times S^2$ mean in geometric topology?

I know the $\#$ symbol refers to a connected sum. So that we delete a disk from each sphere and sew the two spheres according to a map $\phi : S^1 \to S^1$. These could be classified by the corresponding element of the fundamental group $[\phi]=[n]\in \pi_1(S^1)$.

This guess is certainly wrong because the result should be a 4-manifold.

I don't even know which term to put into Google.

cactus314
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It means the connected sum of $n$ copies of $S^2\times S^2$. That is, $\#^2 S^2\times S^2 = S^2\times S^2 \# S^2\times S^2$, $\#^3 S^2\times S^2 = S^2\times S^2\# S^2\times S^2\# S^2\times S^2$ and so on.

Michael Albanese
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It is the $n$-fold connected sum of $S^2\times S^2$, i.e. $$\#^n(S^2\times S^2):=\underbrace{(S^2\times S^2)\#(S^2\times S^2)\#\dots\#(S^2\times S^2)}_\text{$n$-many}.$$ In this particular case, this is just $n$ four-dimensional manifolds glued together along boundaries after removing $4$-balls from the respective manifolds (see this page for examples where we consider $\#^nS^1\times S^1$ for $n=1,2,3$). That is, $\#^nS^2\times S^2$ is a (orientable) 4-manifold with genus $n$.

Sergio Charles
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    Good, though note that $S^2 \times S^2$ is not a torus by any definition; it is a product of spheres, but a torus is specifically a product of some number of circles. It is 4-dimensional, not a surface (which is 2-dimensional). This manifold doesn't really have a better name than $S^2 \times S^2$. –  Aug 27 '18 at 20:06
  • Opss. @MikeMiller – Sergio Charles Aug 27 '18 at 20:07