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Soft Question - Intuition of the meaning of homology groups

I've been studying some homology recently and I know it supposedly counts $n$-dimensional holes. For example, the torus has first homology group $H_1(T^2) = \mathbb{Z} \times \mathbb{Z}$ where one copy of $\mathbb{Z}$ corresponds to the middle hole of the tire and the second copy of $\mathbb{Z}$ corresponds to its hollow inside. $H_2(T^2) = \mathbb{Z}$ where $\mathbb{Z}$ corresponds to the middle hole but considered in $2$ dimensions, I think.

Is that correct? Or what does a $2$-dimensional hole look like?

Then looking at the Klein bottle gives $H_1(K) = \mathbb{Z} \times \mathbb{Z}_2$. This is where my already shallow understanding ends: if there is a copy of $\mathbb{Z}_{n}$ in the homology, what sort of hole is there in the space?

Many thanks for your help.

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marked as duplicate by Grigory M, Qiaochu Yuan Sep 17 '11 at 17:12

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"it counts n-dimensional holes" is simply a slogan, but reality is more complicated than that... In particular, your title question, «what is the exact meaning of homology?», is more or less unanswerable in the sense you want.

I think you should try to consider as many examples as you can, so as to build an intuition about what homology actually counts.

For example, consider the real projective plane $\mathbb P^2(\mathbb R)$. Do you know its fundamental group? Do you see why it is $\mathbb Z_2$. Do you know its $H_1$? Can you find an explicit generator $c$ for this latter group? Can you see why $2c=0$? Can you find a two cycle whose boundary is $2c$?

The torus $T=S^1\times S^1$ has two $1$-dimensional holes. Can you see them? How can you detect them concretely? How can you tell them apart?

When you have exhausted $\mathbb P^2(\mathbb R)$ and $T$, pass on to other spaces. There are many!

 

Also, you ask what a $2$-dimensional hole looks like... Well, do you know what a $1$-dimensional hole looks like? How can you detect a $1$-dimensional hole?

For example, if $L$ is a line in $\mathbb R^3$, then the set $X=\mathbb R^3\setminus L$ has a $1$-dimensional hole. You can detect it in that there are certain vector fields on $X$ which are not the gradient fields of any function $X\to\mathbb R$. One way to see this is to find an example of such a field $F$ such that the circulation of $F$ along a closed curve is not zero.

Now let $P$ be the origin in $\mathbb R^3$. There is no $1$-dimensional hole in $Y=\mathbb R^3\setminus\{P\}$. Can you see why? (as opposed to computing $H_1(Y)$) Also, $Y$ has a $2$-dimensional hole: can you see it? how can you detect its presence?

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I picture a one dimensional hole as something you can not get $S^1$ passed. I'm not sure it's right but in a cell complex such as the torus I can consider the one skeleton and then I see that it contains $2$ one dimensional holes. Now going to look at the other questions you gave me... thank you! –  Matt N. Sep 17 '11 at 7:30
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@Matt: that idea of hole is related to the homotopy group, but in homology the key point is whether curves are boundaries of surfaces or not, and so on. BTW, there other cell decompositions of the torus, which have $1$-skeletons which are much more complicated than the one you are considering: why does that not imply that there are many more $1$-holes? –  Mariano Suárez-Alvarez Sep 17 '11 at 7:31
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