## Homology Primer 3: Introductory Examples!

### January 4, 2011

In this post, we’re going to do something that I thought helped me through homology, but that others no doubt think is a huge waste of time: we’re going to explicitly compute the homology groups of trivial or nearly-trivial things explicitly, without appealing to any theorems regarding homeomorphisms or the like.

Recall the things that we’ve done so far.  The steps to find the homology groups in this post (and, more generally, in life) will be as follows:

1. Find the chain groups and their generators.
2. Compute explicitly the image and kernel of the boundary maps.
3. Find a nice way to express the kernel and image of these boundary maps by potentially using different, but equivalent, generators (we’ll get to this step soon; it’ll make more sense then).
4. Find the quotient $\displaystyle \frac{Ker(\delta_{i-1})}{Im(\delta_{i})}$, which is equal to the $(i-1)$-th cellular homology group.
5. Reflect on this solution: does it make sense?  Is it nice?  Do we love it?

So without further delay, let’s dig right in.

## The 2-dimensional Disk, $D^{2}$.

We can make this disk a number of ways, but let’s do it in the following way.

Define $D^{2}$ by the cell structure which has one 0-cell, one 1-cell, and one 2-cell with the attaching maps defined as below.

We note that the 0-cell is called $v$, the 1-cell is called $e$ and it’s directed as above, and the 2-cell is called $f.$.  Okay.  Now, let’s detail the chain groups.

$C_{0}(D^{2}) = {\mathbb Z}$

Generated by $v$.

$C_{1}(D^{2}) = {\mathbb Z}$

Generated by $e$.

$C_{2}(D^{2}) = {\mathbb Z}$

Generated by $f$.  If these things are mysterious to you, you may want to go back to the previous post and see what these chain groups actually are.  Also note that $C_{i}(D^{2}) = 0$ for every $i > 2$, since we have no cells in that dimension.

Now, let’s consider the boundary maps.  Because these maps are linear transformations, we only need to worry about where they take the generator of our set.  So, for example, we have

$\delta_{2}:C_{2}(D^{2}) \rightarrow C_{1}(D^{2})$

takes our generator of $C_{2}$, which is just $f$, into what?  What is the boundary of this set?  Well, the boundary is just $e$, since this is what we’re gluing the 2-cell to.  Thus, $\delta_{2}(f) = e$ which implies that

$Ker(\delta_{2}) = 0$

$Im(\delta_{2}) = {\mathbb Z}$

where the right hand side is generated by $e$.  Note that the kernel is trivial for a variety of reasons: the first isomorphism theorem, the fact that only 0 maps to 0, and so forth.

Similarly, that $\delta_{1}:C_{1}\rightarrow C_{0}$, so we need only consider $\delta_{1}(e)$.  But what is this?  On one side, we attach $e$ to $v$, and then it wraps around and goes back to $v$.  Thus, we have $\delta_{1}(e) = v - v = 0$.  This means that

$Im(\delta_{1}) = 0$

$Ker(\delta_{1}) = {\mathbb Z}$

Where the latter case is generated by $e$ and is found either by the first isomorphism theorem, or by noting that every element of the chain group goes to $0$.

Last, what is $\delta_{0}$?  It’s kind of silly.  It sends everything to 0.  Thus, we have

$Im(\delta_{0}) = 0$

$Ker(\delta_{0}) = {\mathbb Z}$

Where the latter is generated by $v$.

Okay.  Now, we’ve done the hard part of this.  Now we just modulo things.  First, let’s examine

$\displaystyle \frac{Ker(\delta_{2})}{Im(\delta_{3})}$

Note that the image and kernel of $\delta_{i}$ for any $i$ greater than your largest cell will be trivial.  Thus, this reduces to

$\displaystyle \frac{Ker(\delta_{2})}{0}$

Which we can replace by noting this kernel is just 0, from above.  Thus,

$H_{2}(D^{2}) = 0.$

Okay, good.  Now, examine

$\displaystyle \frac{Ker(\delta_{1})}{Im(\delta_{2})}$

We know that both the kernel and the image are ${\mathbb Z}$ and generated by $e$.  Thus, this quotient is trivial (we’re modding out two groups generated by the exact same elements), and so

$H_{1}(D^{2}) = 0.$

Kind of boring so far.  Okay, let’s do the last one.  Examine

$\displaystyle \frac{Ker(\delta_{0})}{Im(\delta_{1})}$

This reduces to

$\displaystyle \frac{{\mathbb Z}}{0}$

with the top generated by $v$.  This means that

$H_{0}(D^{2}) = {\mathbb Z}.$

Nice!  This is pretty nontrivial!

Let’s reflect on what all this means. Since we have $H_{2}$ is trivial, this means there are no “two dimensional holes.”  We’d expect this for the disk, where there are no holes in general.  Since $H_{1}$ is trivial, we note there are no “one dimensional holes” (like the ones in $S^{1}$), and this makes sense as well.  The last value is kind of weird, but if we think about what “zero dimensional holes” should mean, it translates to the idea that our figure is connected.  Thus, the group $H_{0}$ tells us how many connected components our complex has; it will be one direct sum copy of ${\mathbb Z}$ for each connected component.

## The Circle, $S^{1}$.

This one is a bit easier to represent the cell structure for.  We’re just going to use one 0-cell and one 1-cell in the following way:

So our 1-cell is connected on both ends to our 0-cell.  Now, more briefly, let’s state what we know about this.  If we don’t mention a chain group or a homology group, it’s because it’s trivial.

$C_{1}(S^{1}) = {\mathbb Z}$

Generated by $e$.

$C_{0}(S^{1}) = {\mathbb Z}$

Generated by $v$.

$\delta_{1}(e) = v - v = 0$

$\delta_{0}(v) = 0$

And so we have

$Im(\delta_{1}) = 0$

$Ker(\delta_{1}) = {\mathbb Z}$ generated by $e$,

$Im(\delta_{0}) = 0$

$Ker(\delta_{0}) = {\mathbb Z}$ generated by $v$.

Which gives us the homology groups

$\displaystyle H_{1}(S^{1}) = \frac{Ker(\delta_{1})}{Im(\delta_{2})} = \frac{{\mathbb Z}}{0} = {\mathbb Z}$

$\displaystyle H_{0}(S^{1}) = \frac{Ker(\delta_{0})}{Im(\delta_{1})} = \frac{{\mathbb Z}}{0} = {\mathbb Z}$

Which is a little bit different than what we had before!  Notice that since $H_{1}$ is not trivial (and is generated by $e$, we have that there is a one dimensional hole, and, furthermore, $e$ is the element that surrounds it.  This makes sense, right?  There is a big hole in the middle of this circle, and the element which bounds it is exactly the 1-cell $e$.  The group $H_{0}$ tells us that $S^{1}$ is connected, which is the case.

Parenthetically, showing $S^{1}$ is connected is a good exercise to give to beginner topology students.  Kind of cute.

## $S^{1}$ with a Different Cell Structure.

This time, just to show you that equivalent “looking” cell structures (we’ll talk more about what this means later, but for now homeomorphic will suffice) have the same homology groups, let’s make the circle a different way.

This time, we’ve made it with two 0-cells and two 1-cells.  Let’s compute as in the last example, and, also as in the last example, if we don’t mention a chain group or a homology group, it’s because it’s trivial.

$C_{1}(S^{1}) = {\mathbb Z} \oplus {\mathbb Z}$

Generated by $e_{1}, e_{2}$.

$C_{0}(S^{1}) = {\mathbb Z}\oplus {\mathbb Z}$

Generated by $a, b$.

$\delta_{1}(e_{1}) = b - a$

$\delta_{1}(e_{2}) = a - b$

$\delta_{0}(a) = \delta_{0}(b) = 0$

Remember that this last equality (the boundary of a 0-cell) is null by definition.  Now, we have

$Im(\delta_{1}) = \langle (b-a)\rangle = {\mathbb Z}$

since $a - b = -(b-a)$, we have that this image is just the integers, but it’s generated by the value $b - a$.  That’s kind of strange, but we’ll go with it.  Now think about this: when is the boundary of some 1-chain going to be 0?  If we have

$\delta_{1}(xe_{1} + ye_{2}) = x(b-a) + y(a-b) = (x - y)(a - b) = 0$

Implying that $x = y$.  Thus, we have the set generated by $e_{1} + e_{2}$ is the kernel; all scalar multiples of this chain will give a zero boundary!

$Ker(\delta_{1}) = {\mathbb Z}$ generated by $e_{1} + e_{2}$,

$Im(\delta_{0}) = 0$

$Ker(\delta_{0}) = {\mathbb Z}\oplus {\mathbb Z}$ generated by $a, b$.

But now it’s a little trickier to do the homology groups because the things which generate our sets are not so nice anymore.

$\displaystyle H_{1}(S^{1}) = \frac{Ker(\delta_{1})}{Im(\delta_{2})} = \frac{{\mathbb Z}}{0} = {\mathbb Z}$

Notice that, before, this was generated by our only edge.  Now it’s generated by $e_{1} + e_{2}$ which makes one complete revolution around the circle, disregarding the “second vertex” $b$ in this example.

$\displaystyle H_{0}(S^{1}) = \frac{Ker(\delta_{0})}{Im(\delta_{1})} = \frac{{\mathbb Z}\oplus {\mathbb Z}}{{\mathbb Z}}$

Now, this is not so easy to see.  An inexperienced algebraist might see this and say, oh, of course this is just ${\mathbb Z}$, but we cannot conclude this.  Specifically, we need to look at the generators.  On the top, our generators are $a$ and $b$ for ${\mathbb Z} \oplus {\mathbb Z}$ where each vertex generates each of the direct ${\mathbb Z}$summands.  But what is the bottom?  It’s generated by $b - a$.  Well, this is a bit of a pickle.  The generators don’t really match up nicely.

Well, let’s do something we’re going to be doing a lot.  Let’s try to change the generators on the top to match the generators on the bottom.  We can do this easily if we consider:

$\langle a, b\rangle = \langle b - a, a\rangle$

where these braces mean “set generated by.”  Why is this true?  Given any element of the first one, we can make it using the second one; this is kind of a new way to adapt that linear algebra “change of basis” trick.  Neat.  Okay, now, note that we can make our “numerator” of

$\displaystyle \frac{{\mathbb Z}\oplus {\mathbb Z}}{{\mathbb Z}}$

be generated by $(b-a), a$ and since the bottom is generated by $b-a$, it simply eliminates one of the summands in the top.  This allows us to reduce the calculation to just

$\displaystyle H_{0}(S^{1}) = {\mathbb Z}$.

Note that the argument used here was not an easy one.  We’re doing a lot in these last few steps, so please do this example out and see if you follow what I mean.  It’s a bit difficult to get used to changing the basis of our direct sum to make the quotient nice, and I may do an entirely separate post on this because we use it quite a bit.  Nonetheless, we can note here that we obtain the exact same calculations as we did in the previous section with $S^{1}$.

## Let’s move it up with $S^{2}$.

Surprisingly, $S^{2}$ is really no more difficult than $S^{1}$; in fact, the calculations are almost exactly the same.  Let’s do the “easy” version of $S^{2}$ first by noting that it can be made with a 2-cell attached to a 0-cell.

Here we are letting our 0-cell be $v$ and our 2-cell being $f$.  As usual, let’s write our chain groups down.

$C_{2}(S^{2}) = {\mathbb Z}$

generated by $f$.

$C_{1}(S^{2}) = 0$

since there’s no 1-cells.

$C_{0}(S^{2}) = {\mathbb Z}$

generated by $v$.

Now let’s do our boundary maps!

$\delta_{2}(f) = 0$

as there are no 1-cells for which $f$ is a boundary.  Recall that this boundary map needs to map into the 1-cell chain group, and so it cannot possibly be anything but 0.

$\delta_{1}(0) = 0$

trivially, as there are no 1-cells, and

$\delta_{0}(v) = 0$

by definition.  Now, let’s check the images and kernels.

$Im(\delta_{3}) = 0$, since there are no 3-cells.  (We’ll need this for the calculation of $H_{2}$.)

$Im(\delta_{2}) = 0$ as above.

$Ker(\delta_{2}) = {\mathbb Z}$ generated by $f$ by the first isomorphism theorem.

$Im(\delta_{1}) = 0$,

$Ker(\delta_{1}) = 0$, by first isomorphism theorem,

$Im(\delta_{0}) = 0$,

$Ker(\delta_{0}) = {\mathbb Z}$ generated by $v$ as usual.

Now, let’s calculate homology groups.  It’s a bit easier this time around.

$\displaystyle H_{2}(S^{2}) = \frac{Ker(\delta_{2})}{Im(\delta_{3})} = \frac{{\mathbb Z}}{0} = {\mathbb Z}$

$\displaystyle H_{1}(S^{2}) = \frac{Ker(\delta_{1})}{Im(\delta_{2})} = \frac{0}{0} = 0$

$\displaystyle H_{0}(S^{2}) = \frac{Ker(\delta_{0})}{Im(\delta_{1})} = \frac{{\mathbb Z}}{0} = {\mathbb Z}$

and every other homology group is trivial.  We’d expect something like this, since it states that the sphere is connected (this is what $H_{0}$ is telling us) and that there is a single two-dimensional hole, namely the inside of the sphere (this is what $H_{2}$ is telling us).

Using this intuition regarding the homology groups telling us how many $i$-dimensional holes are in a complex, what do you expect the homology groups to be for two spheres which have one point in common as below?

Hint, we can prove that it is the same as the sphere except in $H_{2}$ where there is an additional copy of ${\mathbb Z}$ (why?  what is it generated by?) and so we have $H_{i} = 0$ for every $i \neq 0, 2$ and $H_{0} = {\mathbb Z}$, $H_{2} = {\mathbb Z}\oplus {\mathbb Z}$.  Calculate this for yourself explicitly.

### 5 Responses to “Homology Primer 3: Introductory Examples!”

1. Landau said

Nice, thanks for writing this up! Little remark: when you say “Now thing about this: when is the boundary of some 1-chain going to be 0? If we have (…) Implying that a=b”, that last bit should be ‘implying that x=y’.

2. matrixbud said

In step 4 at the beginning of the article I believe you meant to put the i+1 in your denominator rather than in the numerator?

Nice article. I learned a lot as I was starting to do such computations myself and wanted some way to check my answers. I learned more than I thought there was to learn. Thank you so much.

Souldn’t the hole alluded to, for the case of the Sphere, and implied by non-vanishing $H_{2}$, be 3-dimensional, instead of 2-dimensional?