# One point compactification of $(\mathbb{C}^{\ast})^n$

I would like to know if there is a closed form formula for the homotopy type of $\widehat{(\mathbb{C^{\ast}})^n}$? For example, it is not difficult to see that $\widehat{\mathbb{C^{\ast}}}$ has the homotopy type of $S^1\vee S^2$.

My guess is that the formula, for general $n$, should look something like this $$\bigvee_{i=0}^n \left(\bigvee_{\binom{n}{i}} S^{n+i}\right).$$ The hunch is based on the fact that the compactly supported cohomology of the $n$-torus has the homology of the above wedge of spheres. However, I am not sure how I can prove this.

Note: I did post the same question on Math StackExchange; here is the link.

• I don't think that what you call "complex torus" is what is usually called "complex torus"... Isn't a complex torus something of the form $\mathbb{C} / L$ for some lattice $L$? (Because when reading your title my first reaction was "they're already compact.) – Najib Idrissi Apr 14 '18 at 8:03
• Yes, I agree and I made changes accordingly. – Priyavrat Deshpande Apr 14 '18 at 15:33
• It is completely standard in Lie theory to call $(\mathbb{C}^*)^n$ a complex $n$-torus. – Dan Ramras Apr 17 '18 at 5:19

If $\widehat X$ is the 1-point compactification of $X$, then there is a homeomorphism (for, say, locally compact Hausdorff spaces) $$\widehat{X \times Y} \cong \widehat X \wedge \widehat Y$$ with the smash product. Moreover, the smash product preserves homotopy equivalences for well-pointed spaces, which $\widehat{\Bbb C^*}$ and $S^1 \vee S^2$ both are.
Therefore, because the smash product distributes over the wedge product, there is a homotopy equivalence \begin{align*} \widehat{(\Bbb C^*)^n} &\simeq (S^1 \vee S^2)^{\wedge n}\\ &\cong \bigvee_{i_k \in \{1,2\}} S^{i_1} \wedge S^{i_2} \wedge \dots \wedge S^{i_n}\\ &\cong \bigvee_{i=0}^{n} \bigvee_{\binom{n}{i}} S^{n+i} \end{align*} as desired.