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Here $\Delta_{\mathbb R^n \times \mathbb R^n}= \{(x,y) \in \mathbb R^n \times \mathbb R^n \mid x=y\}$. My idea was that for $n=1$ we have $$\mathbb R \times \mathbb R \setminus \Delta_{\mathbb R \times \mathbb R} \simeq B^2 \amalg B^2 \simeq 0$$

as the diagonal $\Delta_{\mathbb R \times \mathbb R}$ divides the plane $\mathbb R^2$ into two regions, which are both homotopic to the ball and hence contractible (or we can also argue directly that they are contratible). I suspect that this can be generalized to higher dimension where we have a hyperplane dividing our space into two contractible subspaces. Are there any flaws?

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  • $\begingroup$ At least your argument holds only for $n=1$, because $n$-dimensional hyperplane in $\mathbb{R}^{2n}$ can seperate $\mathbb{R}^{2n}$ into two parts only if $n=2n-1$. $\endgroup$ – cjackal Mar 13 '16 at 14:09
  • $\begingroup$ Hint: In $n=1$ case, what you really do is collapsing one dimension of $\mathbb{R}^2$ in the direction of the diagonal to reduce the problem into calculating the homology(or homotopy type, whatsoever) of $\mathbb{R}\setminus \left\{0\right\}$. Generalize! $\endgroup$ – cjackal Mar 13 '16 at 14:14
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The flaw in your argument for $n>1$ is that the diagonal in $\mathbb R^n\times\mathbb R^n$ is not a hyperplane, but an $n$-dimensional subspace in a $2n$-dimensional euclidean space.

First notice that there is a homeomorphism $\mathbb R^n\times\mathbb R^n\to\mathbb R^n\times\mathbb R^n$ given by $(x,y)\mapsto(x,x-y)$, sending $\Delta$ to $\mathbb R^n \times \{0\}$. Hence, we can study $\mathbb R^n\times \mathbb R^n\setminus(\mathbb R^n\times\{0\})$ instead. Now there is a rather obvious homotopy equivalence $\mathbb R^n\times \mathbb R^n\setminus(\mathbb R^n\times\{0\}) \simeq \mathbb R^n\setminus\{0\}$ and finally $\mathbb R^n\setminus\{0\} \simeq S^{n-1}$.

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Define a map from your space $X = \mathbb{R}^n \times \mathbb{R}^n \setminus \Delta$ to $\mathbb{R}^n$ given by $$ f(\vec{x},\vec{y}) = \vec{x} - \vec{y} $$ and note that this is surjective onto $\mathbb{R}^n \setminus \{0\}$. If you're careful, this will give a good description of $X$ as a fibre bundle. What can you then say about the homology of $X$?

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  • $\begingroup$ We know that $\mathbb R^n \setminus \{0\}$ is homotopic to $S^{n-1}$. Unfortunately I have not encountered fibre bundles yet. $\endgroup$ – noctusraid Mar 13 '16 at 10:40

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