I want to give an elementary example of what de Rham cohomology is about (my own conception of it is not far from this–I don’t have any non-trivial understanding of other things with the word “(co)homology” involved, only some vague ideas like that Stokes’ theorem says that integration exhibits the exterior derivative on differential forms and the boundary operator on manifolds as adjoint, so there should be some dual theory about submanifolds / singular chains / whatever). The example here will just be in terms of vector fields on a plane, so no knowledge of differential forms or manifolds is necessary. Knowing a bit of vector calculus will be helpful, but hopefully not required to get something out of this.
In our example, the idea is to look at what sort of vector fields can live on some two dimensional space to deduce topological information about the space. The relevant properties of vector fields are ones learned in a typical vector calculus class: specifically, the curl of a vector field, and whether or not a vector field is conservative. First of all let’s recall what these are.
The curl of a vector field V is (for our purpose) a function from R² to R which does as the name suggests: the curl of V at a point P is a measure of how much the vector field is rotating about the point P, and in what direction. Imagine that the vector field is describing some sort of force like gravity, or the flow of water, and suppose we place a little circle at P. If the vector field is rotating around P clockwise, it takes some positive amount of work to traverse the circle counterclockwise against the flow of the vector field; if the field is rotating counterclockwise, it does the work for us, and so the amount of work necessary is negative. Of course as the circle gets smaller, so too does the amount of work necessary to traverse it, no matter how much the field is rotating: thus, it’s better to consider the amount of work needed per unit of length. This, then, is the definition of curl: take the amount of work needed to traverse a small circle at P divided by the circumference of the circle, and let the size of the circle go to 0.
For example, for the vector field V = (y, -x), the curl is positive at 0:
curl
In fact this vector field has positive curl everywhere. It takes positive work to go counterclockwise around, say, the circle on the lower right, since the vector field is smaller closer to the origin where it helps us along the circle, and bigger further away where we have to work against it (maybe a better way to imagine curl is as a measure of how much a tiny weather vane at a point will spin due to the vector field). For the record, this vector field V has curl 2 at every point.
Let’s look at a vector field with zero curl. This vector field is defined on R² without the origin; let’s call this space M write W for the vector field. W is given by the formula 
Surely it is outrageous that this vector field W has no curl, it is curling all over the place! But at every point of M, there’s no curl because all of the vectors have length 1, so a weather vane isn’t going to go anywhere, and we can’t put a weather vane at the origin since M is missing the origin!
This vector field provides the first idea of what this example is about. Here we have a vector field which by any decent standards looks like it ought to have nonzero curl, but it fails to, because the space on which it is defined is missing the place where it is trying to curl about! The existence of a vector field like this is giving us topological information about M, namely that it has some kind of hole.
It’s also important that this example does not work on all of R² (we don’t want to be told that R² has a hole in it), because W cannot be defined at the origin in a continuous way: the only choice for a vector there is the zero vector (otherwise which direction would it point in), but this doesn’t work since all of the vectors in the field have length 1. Compare to the first vector field, where the lengths of the vectors decrease to zero in a nice continuous way as you head to the origin. (And certainly we need continuous vector fields to learn anything about the topology of our spaces!)
So this is the first half of the story, getting the idea of looking for vector fields that “should” have nonzero curl but don’t; the second half is deciding how to tell what we mean by “should” here. A vector field is conservative if the amount of work to go from a point P to a point Q along some path does not depend on the path, only on P and Q. A gravitational field, for example, is conservative. Notice that in particular, a conservative vector field has zero curl, because it takes zero work to go around any circle: starting at P and ending up at P via a circle must take the same amount of work as just sitting at P for a while.
Notice also that the vector field W is not conservative: it takes positive work to go counterclockwise in a circle around the origin. This is what we mean by vector fields which “should” have nonzero curl but don’t: vector fields that have zero curl but are not conservative. For an open subset U of R² (like R² with the origin removed), write 
The first de Rham cohomology group of U, which we will write 
This pretty much concludes the example. As the terminology suggests, there are higher de Rham groups (one for every nonnegative integer), which might be thought of as detecting higher-dimensional holes. We can also talk about the de Rham cohomology of any smooth manifold: the dimensions of the de Rham cohomology groups of the 2-sphere are 1, 0, 1, 0, 0, 0, … The dimension of the 0th de Rham group is the number of connected components of the manifold, and the remaining numbers fit the idea that the 2-sphere has one “two-dimensional hole”, and no holes of other dimensions.
It is important, and easy to believe, that de Rham cohomology is a diffeomorphism invariant: two diffeomorphic smooth manifolds have the same de Rham cohomology. In fact it’s a homeomorphism invariant, but even better, it’s a homotopy invariant: if one manifold can be continuously deformed into another, the two manifolds have the same de Rham cohomology. For example, R² with the origin removed is homotopic to a circle, via the homotopy that stretches everything inside the unit circle up to it, and everything outside the unit circle down to it, so the circle also has first de Rham cohomology group of dimension 1 (as it should). Thus, the de Rham cohomology can be used as a tool to decide whether two manifolds are diffeomorphic / homeomorphic / homotopy invariant.
, and any matrix A with 
will fix the ellipse.
for the usual Euclidean dot product (in fact any inner product on 
will do here). Define a new inner product 
, either thinking of this integral as over G as a compact subset of 
, or by choosing a volume form on the Lie group G. If A is in G, the change of variables formula immediately gives 
since A preserves the volume of G, or because G is compact).
; since B arises from an inner product, it is symmetric and positive-definite. We want to use B to find M invertible such that 
is orthogonal when A is G, i.e. 
. But we know that 
for all 
, so any M with 
will do.
is diagonal; since B is positive definite, the diagonal entries of D are all positive, and so D has a square root 
, gotten by taking the square root of the entries of D. Then 
is symmetric since U is orthogonal, and 
.
is a subgroup of O(n). At this point let’s go back to the plane. Since H is infinite and compact, it cannot have dimension 0 as a Lie group. O(2) has dimension 1, so G has dimension 1, and therefore must be a union of connected components of O(2) since it is an embedded submanifold. The connected components of O(2) are SO(2) and the set of reflections, and so the possibilities for H are O(2) and SO(2).
is exactly the symmetry group of 
. Take v in 
where U is orthogonal and 
is evidently an ellipse; the disk S gets scaled along each axis by the corresponding diagonal entry of 
with the usual Euclidean distance, so we want to check that G is closed and bounded.
is a sequence in G converging to A in GL(2,R), then for any p in K, 
converges to Ap by continuity. Since K is closed, Ap lies in K, so 
. Applying the same argument to 
gets AK = K, so A is in G.
. The claim is that given R > 0, there is M such that |A| > M implies |Av| or |Aw| > R; here |A| is just the Euclidean norm thinking of A as a member of 
for all A in G.
. Then 
are just the columns of A, so 
. Thus, if 
, then either |Ae_1| or |Ae_2| must be bigger than R.
. By the above, if 
, then one of 
or 
is bigger than R. But 
, so taking 
forces |Av| or |Aw| > R.
for the area (e.g. Lebesgue measure) of K. Let 
be the maps 
and 
. The centroid of K is the point 
. There’s one problem here: what if 
? Wikipedia suggests the definition “the intersection of all straight lines that divide X into two parts of equal moment about the line”, which seems like kind of a pain to work with. Luckily we’re talking about convex sets here. If K is convex and isn’t a line segment (we don’t care about those anyway), then it has positive measure: given three non-collinear points in K, the solid triangle with those vertices must lie in K, and that has positive measure.
, and write T = v + A, where A is linear. Because T(K) has area 
, we must have 
. Using the change of variables formula and the relation A(K) = K – v, the right side is 
. Changing variables again gives ![v + \frac{1}{\mu(K)}\left[(\int_K \pi_1, \int_K \pi_2) - \mu(K)v\right]](http://s0.wp.com/latex.php?latex=v+%2B+%5Cfrac%7B1%7D%7B%5Cmu%28K%29%7D%5Cleft%5B%28%5Cint_K+%5Cpi_1%2C+%5Cint_K+%5Cpi_2%29+-+%5Cmu%28K%29v%5Cright%5D+&bg=ffffff&fg=333333&s=0 "v + \frac{1}{\mu(K)}\left[(\int_K \pi_1, \int_K \pi_2) - \mu(K)v\right]")
, and this is exactly 
.![[-1,1] \times \{0\}](http://s0.wp.com/latex.php?latex=%5B-1%2C1%5D+%5Ctimes+%5C%7B0%5C%7D&bg=ffffff&fg=333333&s=0 "[-1,1] \times \{0\}")
is fixed by any linear map with matrix 
for 
. What we’ll show is that these are the only counterexamples: any compact convex set K fixed by infinitely many members of AGL(2,R) is a line segment or an ellipse.
such that 
for all x, y. This is all completely standard, but it’s good to know. We don’t demand any nice properties to start with, although many follow from the definition. Suppose T is an isometry of 
, so the condition that T is an isometry is equivalent to the condition that 
, i.e. that T preserves the angles between vectors. Since 
, and this must equal 
for all x, T being an isometry means that 
is the identity map. That is, if we think in terms of matrices (with respect to the standard basis), the group of linear isometries of 
.
, and it’s helpful to distinguish these: the matrices in O(n) of determinant 1 form a subgroup, SO(n). At this point let’s restrict to the plane. It’s just a matter of algebra to check that every member of SO(2) looks like 
where 
. The latter condition means we can write this more suggestively as 
. This linear map sends (1,0) to 
and (0,1) to 
: it’s exactly the counterclockwise rotation by 
.
. One can check that this matrix fixes 
and sends the perpendicular vector 
to its negative. That is, everything in O(2) not in SO(2) is a reflection across a line through the origin.
moves right. The answer for general n turns out to be a little bit surprising, I think.
. Directly counting the number of Dyck words beginning with a specified number of R’s is a little tricky. The obvious thing to do is forget the R’s and hope that what’s left is something nice that we can count. Of course we do always get an (n-k, n)-bad word, since the result always starts with U, but we can’t get all (n-k, n)-bad words like this, or even all of them that start with U: in the example with n = 3, k = 2, removing the initial R’s from RRURUU and RRUURU gives us URUU and UURU, but UUUR is nowhere to be found, since RRUUUR isn’t a Dyck word.
.
(1)
and 
, then
(2)
coefficient of f(x)g(x), namely 
, looks kind of like the sum 
from (1). And indeed, if we set
,
from (1).
and multiplying by x, twice, shows that
,
, the binomial theorem gives 
. Now nD(n,n)A(n) is the 
, and we can use the binomial theorem to see that this coefficient is 
. Thus, using the formula for D(n,n) we found before,![A(n) = \frac{n+1}{n}{2n \choose n}^{-1} \left[{2n+1 \choose n+2} + {2n \choose n+2}\right].](http://s0.wp.com/latex.php?latex=A%28n%29+%3D+%5Cfrac%7Bn%2B1%7D%7Bn%7D%7B2n+%5Cchoose+n%7D%5E%7B-1%7D+%5Cleft%5B%7B2n%2B1+%5Cchoose+n%2B2%7D+%2B+%7B2n+%5Cchoose+n%2B2%7D%5Cright%5D.&bg=ffffff&fg=333333&s=0 "A(n) = \frac{n+1}{n}{2n \choose n}^{-1} \left[{2n+1 \choose n+2} + {2n \choose n+2}\right].")
.
: for large n, an (n,n)-Dyck path begins with about 3 moves right. Too much shouldn’t be read into it, though; there are always going to be more Dyck paths that start with 1 R (or 2 R’s) than that start with 3, it’s just that the average happens to be about 3 for large n. It’s worth noting that the fraction of Dyck paths starting with k R’s also has a nonzero limit for each k, e.g. for k = 1 and k = 2 it’s 1/4, and for k = 3 it’s 3/16.
(otherwise there aren’t any). But then n-1 < m+1, which means that every (n-1, m+1)-word is a bad word.
ways to choose which positions the R’s go in for such a word, and then the positions of the U’s is completely determined, so the number of (n-1,m+1)-words is 
, and so the number of (m,n)-Dyck words (paths) is the difference of these numbers,
, which start (going from n = 1): 1, 2, 5, 14, 42, …, are called the Catalan numbers, and are deservedly famous, since they count everything. D(n,n) is, for example, the number of strings of 2n brackets which are paired correctly, so that D(3,3) = 5 corresponds to the strings [[[]]], [[][]], [[]][], [][[]], [][][]. Or, D(n,n) is the number of ways to cut a regular (n+2)-gon into triangles by joining the vertices with lines. And so on. Wikipedia has 
in base 10. Replace it with 
. Eventually you’ll get to 2997.
.
, let 
denote the result of the rule applied to 
, so that eventually the rule must lead to some small multiples of 999 that can be checked by hand.
for the sum of the digits of 
. The most notable properties of the digit sum function 
in base 10; since 
and 
, we get that
and 
(resp. 9) iff 
(resp. 9).
. Thus, by the previous sentence, 3 divides 
, and so 9 divides 
. Similarly, this means that 9 divides 
, so 999 divides 
. So, after at most 3 applications of the rule, we get a multiple of 999.
.
, so that 
. One can solve 
numerically and find that 
. It’s pretty clear thinking of graphs that 
will then hold for 
, and we can be more precise by looking at derivatives: 
is decreasing, and 
. Thus 
for 
