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Is there a chain representing a generator of $H_q(S^q)$ consisting of only one simplex?

Here $S^q$ denotes the q-dimensional sphere and the answer can be given either for simplicial or singular homology.

The question makes sense only for q > 0, since $H_0(S^0) \cong \mathbb{Z} \oplus \mathbb{Z}$.

I know already that a generator can always be represented by a chain consisting of two simplices, e.g., gluing two q-simplices $\Delta_1^q, \Delta_2^q$ along their boundary in such a way that their difference $\Delta_1^q - \Delta_2^q$ is a cycle. Then this cycle represents also a generator of $H_q(S^q)$.

In the case q = 1, the answer is yes, since we can wrap the unit interval once around the circle such that the two endpoints of the interval are mapped to one point on the circle. So we get a cycle and one can also show that this cycle is homologous to $\Delta_1^1 - \Delta_2^1$, which is a generator of $H_1(S^1)$.

But what about the case q > 1?

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As Martin points out below, this cannot be done in general in singular (simplicial) homology, but it can definitely be done using cellular homology. The sphere is a cell complex with a 0-cell and an n-cell, and the cellular homology chain complex has one generator for every cell. –  Grumpy Parsnip Feb 7 '12 at 17:07

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I think that, for odd dimensional spheres, you might be able to proceed in the same way as you did for $\mathbb{S}^1$. Any $n$-simplex is homeomorphic to an $n$-cell (or $n$-ball), and gluing the boundary of an $n$-cell to a single point, you get the $n$-sphere. If $n$ is odd, the singular $n$-simplex $\varphi$ obtained this way is a cycle; we have $$ \partial \varphi = \sum_{i = 0}^{n} (-1)^{i} \partial_{i}\varphi, $$ and since $\partial_{0} \varphi = \partial_{1} \varphi = \ldots = \partial_{n} \varphi$, and the number of terms is even, we get $0$ (for $n$ even, we have $\partial \varphi = \partial_0 \varphi \neq 0$). I don't think it is going to be too difficult to show that this simplex is homologous to the generator $\Delta_1^n - \Delta_2^n$.

For the even case, you actually need two simplices; no singular $2k$-simplex $\psi$ for $k \geq 1$ can ever be a cycle, since $\partial \psi$ is an alternating sum of an odd number of basis elements in a free abelian group.

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Thank you very much! –  AlexE Feb 7 '12 at 17:04

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