# Euler Characteristic and cellular homology of $\ell^{th}-$space suspension.

I am learning the Euler characteristic in terms of cellular homology and number of cells. Namely, for a space $$X$$, we have $$\sum_{j}(-1)^{j}c_{j}(X)=\chi(X)=\sum_{j}(-1)^{j}rank(H_{j}(X)),$$ where $$c_{j}(X)$$ is the number of $$j$$-cells in $$X$$.

The only note here, https://pages.uoregon.edu/ddugger/hw634-1.pdf, the Q5 of it, seems to claim that there is a way to connect $$\chi(\Sigma^{\ell}X)$$ with $$\chi(X)$$, where $$\Sigma^{\ell}X$$ is the $$\ell^{th}$$ suspension of $$X$$.

I tried to prove it but failed: since we have $$\chi(\Sigma^{\ell}Y)=\sum_{j}(-1)^{j}rank(H_{j}(\Sigma^{\ell}Y))$$ the only thing we need to do is to compute $$H_{j}(\Sigma^{\ell}Y)$$.

To do so, the only thing we can think about is the reduced homology: $$\overline{H}_{j}(\Sigma^{\ell}Y)=\left\{ \begin{array}{ll} H_{j}(\Sigma^{\ell}Y),\ \ \ \text{for}\ j>0;\\ \mathbb{Z}^{\{\#\ \text{of path components} - 1\}},\ \ \ \text{for}\ j=0. \end{array} \right.$$

But by the well-known Suspension Isomorphism, we have $$\overline{H}_{j}(\Sigma^{\ell}Y)=\overline{H}_{j-\ell}(Y).$$

But then what should I do next? Or is there another easier way to compute?

Thank you!

• So you have figured out the homology of the suspension and know that the Euler characteristic is the alternating sum of the homologies, and you are trying to figure out the Euler characteristic of the suspension? Commented May 19, 2020 at 16:07
• @ConnorMalin Yes, I know Euler characteristic is the alternating sum of the rank of the homologies. I am not quite figuring out the homology of suspension. Since, if don't know how to compute further starting from $$\overline{H}_{j}(\Sigma^{\ell}Y)=\overline{H}_{j-\ell}(Y).$$ Yes, I am trying to figure out the Euler charateristic of the suspension. Commented May 19, 2020 at 16:09
• How about you start by defining a "reduced Euler characteristic" by the alternating (starting with a negative) sum of the nonzero homologies. Then it should be clear by your isomorphism how to relate the "reduced Euler characteristics" of a space with its l-fold suspension. Then figure out how to relate the "reduced Euler characteristic" with the Euler characteristic (hint: subtract). Commented May 19, 2020 at 16:11
• @ConnorMalin so you mean starting by $$\sum_{j}(-1)^{j}rank(\overline{H}_{j}(\Sigma Y))?$$ Commented May 19, 2020 at 16:13
• The formula you wrote down for the reduced homology is wrong: in degree $0$ you should have $\mathbb{Z}^{\text{# of path components}} - 1$. Commented May 19, 2020 at 16:19

As Connor Malin suggests we could define a "reduced Euler Characteristic"

$$\tilde{\chi}(X) = \sum_i (-1)^irank(\tilde{H}_i(X))$$

and then notice two things:

1) $$\chi(X) = 1 + \tilde{\chi}(X)$$,

2) by the suspension isomorphism $$\tilde{\chi}(X) = -\tilde{\chi}(\Sigma X)$$.

Putting these together we have

\begin{align}\chi(\Sigma X) &= 1 + \tilde{\chi}(\Sigma X) \\ &= 1 - \tilde{\chi}(X) \\ &= 1 - (\chi(X) - 1) \\ &= 2 - \chi(X).\end{align}

It follows that $$\chi (\Sigma^{2k}X) = \chi(X)$$ and $$\chi(\Sigma^{2k+1}X) = 2 - \chi(X)$$. As a "sanity check" recall that $$\chi(S^{2k}) = 2$$ and $$\chi(S^{2k+1}) = 0$$.

• so we have $$\chi(X)-1=\tilde{\chi}(X)=-\tilde{\chi}(\Sigma X)=-\chi(\Sigma X)+1$$ so that $$-\chi(\Sigma X)=\chi(X)-2?$$ Commented May 19, 2020 at 16:28
• Yes that's right, or if you rearrange you get $\chi(\Sigma X) = 2 - \chi(X)$ (or equivalently $\chi(X) = 2 - \chi(\Sigma X)$) Commented May 19, 2020 at 16:34
• Yes you were right about I made a sign error. So by the same idea, we have $$\chi(X)-1=\tilde{\chi}(X)=(-1)^{\ell}\tilde{\chi}(\Sigma^{\ell}X)=(-1)^{\ell}(\chi(\Sigma^{\ell}X)-1),$$ and thus $$\chi(X)=(-1)^{\ell}\chi(\Sigma^{\ell}X)+(-1)^{\ell+1}+1,$$ using $\ell=2$ do matches you $\ell=2$ example. Am I right? Commented May 19, 2020 at 16:37
• Yes that looks right. In fact there are only two possible values, depending on the parity of the suspension: $\chi(\Sigma^{2k} X) = \chi(X)$ and $\chi(\Sigma^{2k + 1}X) = 2 - \chi(X)$. Commented May 19, 2020 at 16:39
• Yes you are right. Thank you so much! Commented May 19, 2020 at 16:41