I am learning about simplicial homology from Hatcher's Algebraic Topology book. He has examples of some simplicial homology computations for familiar/simple spaces (the torus, circle, and $\mathbb{R} \mathbb{P} ^2$). However, I am interested in doing a computation which involves finding a triangulation of something "more complicated," such as $S^1 \vee S^2$, and over $\mathbb{Q}$ instead of $\mathbb{Z}$ (what does the triangulation of $S^1 \vee S^2$ even look like?). That is, what is a triangulation of $S^1 \vee S^2$ and how can we compute the simplicial homology of $S^1 \vee S^2$ over $\mathbb{Q}$? (If this has been done before, I would appreciate a reference as I have tried searching for this online, and I apologize in advance if this is a duplicate.)
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2Generally, if you want to find the triangulation of $M_1 \lor M_2$, you find the triangulation of $M_1$ and $M_2$ individually and then "combine" them by giving the same label to the distinguished 0-simplex of each of their triangulations. – Mark Saving Apr 20 '21 at 20:08
1 Answers
As long as one knows a triangulation of $\mathbb{S}^{1}$ and a triangulation of $\mathbb{S}^{2}$, one knows a triangulation of the wedge sum (as is probably clear) where one joins both spaces at the point [0]. $\mathbb{S}^{2}$ is homeomorphic to the boundary of a 3-simplex, by embedding the boundary of the 3-simplex in such a way that one encloses the origin in $\mathbb{R}^{3}$, and doing $x \mapsto \frac{x}{|x|}$. It is easy to check this is a bijective and continuous map of compact Hausdorff spaces, and thus a homeomorphism. Similar logic works for $\mathbb{S}^{1}$, and indeed for all higher dimensional spheres. This gets you free triangulations from the n-simplex.
Computing homology with rational coefficients can be done by taking the free $\mathbb{Q}$ vector space on the simplices in each dimension of this triangulation and computing simplicial homology as you would with $\mathbb{Z}$. As you might notice, this is the same operation as tensoring the standard $\mathbb{Z}$-coefficient simplicial chain complex with $\mathbb{Q}$, and in general this is how one computes any homology with coefficients. $\mathbb{Q}$ is flat over $\mathbb{Z}$ and a little work with flatness will show you that the computation of these homology groups agrees with $H_{n}(X; \mathbb{Z})\otimes_\mathbb{Z} \mathbb{Q}$. This does not work at finite fields, as these are not flat over $\mathbb{Z}$, and one usually gets an extra $\operatorname{Tor}$ term showing up in $H_{n}(X;\mathbb{F}_{p})$; this is dictated by something called the Universal Coefficient Theorem. This theorem is mostly likely the reference you are looking for, and is both easily searchable online and in Hatcher!
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You'll also learn about cellular homology later on in Hatcher, which is a much more tractable way to compute homology of CW-complexes, which spheres are the prototypical example of. – Aragogh Apr 20 '21 at 20:23
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Does the final result/computation result with direct sums of copies of $\mathbb{Z}$ then? The person who asked the question probably understands this reasoning, but I was curious in your explanation not knowing about flats and tensors. – Dominated Convergence Theorem Apr 20 '21 at 20:36
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Well the final simplicial homology of $\mathbb{S}^{1} \vee \mathbb{S}^{2}$ is one copy of $\mathbb{Z}$ in degrees 0,1,2, you can check this directly using the triangulation I gave or another one of your choice. The tensor products stuff was only for doing computations over $\mathbb{Q}$. – Aragogh Apr 20 '21 at 21:10