Abstract Nonsense

Singular Homology (Pt. II)

Point of Post: This is a continuation of this post.

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Ok, so we have our notion of $n$-simplexes on general topological spaces. The next key is this, we’ve said that the boundaries of $n$-simplexes are “sums” of $n-1$-simplexes. What does sum mean? Well, we can obviously “force” the notion of “sum” by defining for each $n\in\mathbb{N}$ the singular $n$-chains on  $X$ with coefficients in $R$ to be the free left $R$-module on the set of all singular $n$-simplexes on $X$, we shall denote this module $\Delta_n(X;R)$. Why is this useful? Well, to begin with it gives us a convenient way to describe these “closed objects” we have been discussing. For example, the standard $1$-simplex is merely a line, identifiable with the unit interval $I=[0,1]$. We see then that a singular $1$-simplex on $X$ is nothing more than a path $\sigma:I\to X$.  But, for paths we have an obvious intuition for what it means to be closed–namely $\sigma$ should be closed if $\sigma(0)=\sigma(1)$. Ah! But, this condition can be phrased quite nicely by saying that $\sigma(1)-\sigma(0)=0$ if we are thinking of $\sigma(0),\sigma(1)$ as living inside of $\Delta_0(X)$.  More generally, it allows us define when an object which is the union of $n$-simplices is “closed”. For example, when is the union of three $1$-simplices $\sigma_1,\sigma_2,\sigma_3:I\to X$ closed? Well we have the endpoints $\sigma_i(0),\sigma_i(1)$ for $i=1,2,3$. Intuitively it should be closed if we have that $\displaystyle \sum_{i=1}^3 (\sigma_i(1)-\sigma_i(0))=0$. Indeed, by a quick case analysis one can check that this is true if and only if one of the following three situations holds: the union of the three looks like a triangle (this is the case $\sigma_i(1)=\sigma_{i+1}(0)$ where the indices are modulo three), the union is the union of a loop formed by bending and connecting two of the line segments at the endpoints and the union of a closed loop (this corresponds (up to a rearrangement of the vertices) to the cases $\sigma_1(0)=\sigma_2(0)$, $\sigma_1(1)=\sigma_2(1)$, $\sigma_3(0)=\sigma_3(1)$ and $\sigma_1(0)=\sigma_2(1),\sigma_1(1)=\sigma_2(0),\sigma_3(0)=\sigma_3(1)$), and to the case that we have the union of three closed loops (which corresponds to the case $\sigma_i(0)=\sigma_i(1)$ for $i=1,2,3$). Moreover, it’s clear that the union of simplices can form a “closed region” (in the intuitive sense) if and only if one of these cases holds.

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More generally, one can easily convince one’s self that the union of $m$ $n$-simplices $\sigma_i:I\to X$ where $i=1,\cdots,m$ is going to be closed in the intuitive sense if and only if $\displaystyle \sum_{i=1}^{m}(\sigma_i(1)-\sigma_i(0))=0$. But! We can rephrase all of this by saying that we have a linear map $\Delta_1(X;R)\to\Delta_0(X;R)$ defined on the basis of $1$-simplices by $\displaystyle \sigma\mapsto \sigma(1)-\sigma(0)$ and thus the closed union of simplices can be thought of as precisely those in the kernel of this map! So, we are definitely getting some mileage in phrasing things algebraically. Of course, it takes a little more thought but we can probably convince ourselves that in the general dimension the closed $n$-forms should be precisely those in the kernel of the map $\Delta_n(X)\to\Delta_{n-1}(X)$ given on basis vectors by $\displaystyle \sigma\mapsto \sum_{j=0}^n (-1)^j\sigma_{\mid [e_0,\cdots,\widehat{e_j},\cdots,e_n]}$ where the hat denotes the omission of that vector from the convex span. In other words, the summands are signed images of the normal faces of the $n$-simplex under the map $\sigma$. These are called the $n^{\text{th}}$ boundary maps and are denoted $\partial_n$.

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Remark: The notation $\partial_n$ has to do with the fact that, as we shall see, we are naturally going to create a “differential” complex of $R$-modules–the $\partial$ is supposed to be a messed up $d$.

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The name for the boundary maps are suggestive, and rightfully so. Let’s see if we can convince ourselves that the boundary of a singular $n$-simplex should be the image of that $n$-simplex under $\partial_n$. It’s obvious that if we interpret the sum operation in $\Delta_n(X)$ as signed unions that at least the union is correct–it’s just the union of the faces, or the images thereof. The question is to why the signs should work out the way they do. Start with the case of a singular $2$-simplex–in fact, to make things simple let’s deal with the standard $2$-simplex $\Delta^2$. So, intuitively $\partial_2(\Delta^2)$ should be a “signed” union of lines, namely those forming it’s boundary: $[e_0,e_1],[e_1,e_2],[e_0,e_2]$. The question then is why should the signed union be of the form $[e_0,e_1]+[e_1,e_2]-[e_0,e_2]$? For us we’d like to think about the unions as being “directed”, in the sense that they form a path carried out in time. So, if we fix $e_0$ as our starting vertex then we see that the natural path that makes up the boundary of $\Delta^2$ is $e_0\to e_1\to e_2\to e_0$ corresponding to the sum $[e_0,e_1]+[e_1,e_2]+[e_2,e_0]$. Great. The problem is that, to normalize things, we’d like to express segments in ascending order of  indices. The question then is how $[e_2,e_0]$ relates to $[e_0,e_2]$? Well, intuitively if we think about the path $[e_0,e_2]+[e_2,e_0]$ we see that we end up going nowhere so that this path should be “zero” and so $[e_2,e_0]=-[e_0,e_2]$. But, we then see that the boundary of $\Delta^2$ should be $[e_1,e_2]-[e_0,e_2]+[e_0,e_1]$ as desired. Playing around with the higher-dimensional analogues of making a tour of the vertices will convince you that we have indeed defined the signage of the unions correctly.

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Intuitively then we should have that $\text{im }\partial_n\subseteq\ker\partial_{n-1}$ since, after all, the boundaries higher-dimensional simplices are closed. But, we still need to verify this. Of course this is equivalent to saying that $\partial_{n-1}\circ\partial_n=0$. To check this we merely brute force it

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\displaystyle \begin{aligned}\partial_{n-1}(\partial_n(\sigma)) &= \sum_{j=0}^{n}(-1)^j\partial_{n-1}(\sigma_{[e_0,\cdots,\widehat{e_j},\cdots,e_n]})\\ &= \sum_{j=0}^{n}\left(\sum_{i=0}^{j-1}(-1)^j\sigma_{[e_0,\cdots,\widehat{e_i},\cdots,\widehat{e_j},\cdots,e_n]}+\sum_{i=j+1}^{n}(-1)^{j+1}\sigma_{[e_0,\cdots,\widehat{e_j},\cdots,\widehat{e_i},\cdots,e_n]}\right)\end{aligned}

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The key now is to observe that each of the terms $\sigma_{[e_0,\cdots,\widehat{e_r},\cdots,\widehat{e_s},\cdots,e_n]}$ shows up exactly twice in this sum, once with a positive sign and once with a negative sign. Consequently, the sum is zero. Since $\partial_{n-1}\circ\partial_n$ annihilates a basis for $\Delta_n(X;R)$ it must be the zero map, proving our claim. Putting this all in one concise form we note that we have the following chain of maps

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$\cdots\to\Delta_3(X;R)\overset{\partial_3}{\longrightarrow}\Delta_2(X;R)\overset{\partial_2}{\longrightarrow}\Delta_1(X;R)\overset{\partial_1}{\longrightarrow}\Delta_0(X;R)\overset{\partial_0}{\longrightarrow}0$

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where the composition of two adjacent maps is zero.

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It’s about time we name the images and kernels of the boundary maps since we shall obviously be using them so often. We call the image of the map $\partial_{n+1}$ (unimaginatively) the $n$-boundaries and denote them $B_n(X;R)$ and we call the kernel $\ker\partial_n$ the $n$-cycles (cycle being a generic word for something that encloses an area) and denote it $Z_n(X;R)$. The above statement then amounts to the fact that $Z_n(X;R)\supseteq B_n(X;R)$.

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We in a great position now. We have formally defined the objects that “should” (in a perfect space, with no holes) be the boundaries of higher dimensional objects, the $n$-cycles, and we have formally defined boundaries of higher dimensional objects, the $n$-boundaries. Our intuition then is that a space should have no $n$-dimensional holes if these two sets coincide. But, as we well know, the world is not perfect. We cannot reasonably expect that all spaces have holes, in fact, if this were true all of this would be moot. Moreover, we should hope to get more out of all of this then just the binary answer to “Does the space have holes?” We would like to, in the event of there existing holes, how many there are, or perhaps better put, to what extent does the statement “Does the space have holes?” fail. We can accomplish this in a fairly obvious way. Algebraically we are asking to what extent does $B_n(X;R)$ fail to be $Z_n(X;R)$, and there is an obvious algebraic way to measure this–the quotient. Indeed, define the $n^{\text{th}}$ homology module of $X$ with coefficients in $R$, denoted $H_n(X;R)$, to be the quotient $Z_n(X;R)/B_n(X;R)$. Then, we see that our space will have “no $n$-dimensional holes” when $H_n(X;R)=\{0\}$ and how “large” $H_n(X;R)$ is should be an indication of how badly the space has holes.

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So, there we have it. We have now created a way to algebraically measure “holieness” of a space, which is precisely what we have set out to do. We still have yet to prove that this notion should be a topological invariant, but we shall save this for a general nonsense argument, the motivation of which will be precisely that fact!

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References:

[1] Rotman, Joseph J. An Introduction to Algebraic Topology. New York: Springer-Verlag, 1988. Print.

[2] Bredon, Glen E. Topology and Geometry. New York: Springer-Verlag, 1993. Print.

[3] Hatcher, Allen. Algebraic Topology. Cambridge: Cambridge UP, 2002. Print.

February 21, 2012 -