# Action of symmetry group on homology?

For a singular n-simplex $$\alpha$$ and a permutation $$t\in S_{n}$$, define $$t\alpha$$ to be the simplex with vertices permuted by $$t$$. Do we have $$t\alpha$$ homologous to $$\text{sgn}(t)\alpha$$ ? And does this gives an action of $$S_n$$ on $$H_n$$?

I can see for $$n=1$$ this is true, since the reverse of vertices gives opposite path in fundamental group, and $$H_1$$ is the abelianization of it.

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Nov 13, 2023 at 6:44
• Commented Jul 30 at 14:04

A problem with this construction is that the vertex permutation operation does not necessarily map cycles to cycles. Therefore, it cannot possibly define a map on homology groups.

For example, let's choose $$X = \mathbb R^3$$ as our topological space, and let $$p, q, r, s$$ be any four distinct points in $$X$$. Let $$\tau \in C_2(X)$$ be the chain $$\tau = [q, r, s] - [p, r, s] + [p, q, s] - [p, q, r],$$ where, for example, the singular $$2$$-simplex $$[q, r, s]$$ is defined to be the affine map from the standard $$2$$-simplex to the triangle with vertices $$q, r, s$$.

Applying the boundary map $$\partial$$ to $$\tau$$, we obtain \begin{align} \partial (\tau) & = [r, s] - [q, s] + [q, r] - [r, s] + [p, s] - [p, r] + [q, s] - [p, s] + [p, q] - [q, r] + [p, r] - [p, q] \\ &= 0. \end{align} This shows that $$\tau$$ is a cycle.

Now, take $$t \in S_3$$ to be the permutation $$t = (1, 2, 3)$$. Acting on $$\tau$$ with $$t$$ by vertex permutation, we obtain $$t \tau = [r, s, q] - [r, s, p] + [q, s, p] - [q, r, p].$$

And now, if we apply the boundary map $$\partial$$ to $$t \tau$$, we get \begin{align} \partial (t \tau) & = [s, q] - [r, q] + [r, s] - [s, p] + [r, p] - [r, s] + [s, p] - [q, p] + [q, s] - [r, p] + [q, p] - [q, r] \\ & = [s, q] + [q, s] - [r, q] - [q, r], \end{align} which does not vanish. Thus $$t \tau$$ is not a cycle.

If $$\alpha$$ is a singular $$n$$-simplex, then is $$t\alpha$$ homologous to $$\alpha$$?

In other words:

If $$\alpha$$ is a singular $$n$$-simplex, then does there exist a singular $$(n+1)$$-chain $$\rho$$ such that $$t\alpha - \alpha = \partial \rho$$?

Suppose that the statement is true for all singular $$n$$-simplexes $$\alpha$$. Then the statement must be true for all singular $$n$$-chains as well.

Let's consider the example of the singular $$2$$-chain $$\tau$$ on $$X = \mathbb R^3$$, as defined above. Notice that $$\tau = \partial ([p, q, r, s]).$$

Suppose there exists a singular $$3$$-chain $$\rho$$ such that $$t\tau - \tau = \partial \rho.$$

Then it would follow that $$t\tau = \partial([p, q, r, s] + \rho).$$

Therefore, $$\partial(t\tau) =\partial \partial([p, q, r, s] + \rho) = 0,$$ since the boundary of a boundary is always zero.

This contradicts our earlier calculation, where we found that $$\partial(t\tau)$$ is non-zero.

• In page 211 of Hatcher’s algebraic topology, he proved that the action of $t$ that reversing the order of vertices of simplex with a sign modification is homotopic to identity, does this mean that, in this special case, two problems have positive answer? Commented Nov 15, 2023 at 6:42
• For the two problems you posed to have a positive answer, the action of of $t$ should be a chain map, i.e. $t\partial=\partial t$. If $t$ is a chain map, then $t$ maps cycles to cycles and boundaries to boundaries. Hence $t$ defines s map on homology. Commented Nov 15, 2023 at 9:03
• Being homotopic to the identity is something you try to prove after you have already shown that $t$ is a chain map. Being homotopic to the identity implies that the action of $t$ on homology is the identity homomorphism. Commented Nov 15, 2023 at 9:05