# Reading Merkulov: Differential geometry for an algebraist (1 in a series)

Published: Sat 11 February 2006

I'll do this in posts and not pages on further thought...

Sergei Merkulov at Stockholm University gives during the spring 2006 a course in differential geometry, geared towards the algebra graduate students at the department. The course was planned while I was still there, and so I follow it from afar, reading the lecture noteshe produces.

At this page, which will be updated as I progress, I will establish my own set of notes, sketching at the definitions and examples Merkulov brings, and working out the steps he omits.

## Familiar parts in unfamiliar language

Merkulov begins the paper by introducing in swift terms the familiar definitions from topology of topology, continuity, homeomorphisms, homotopy, and then goes on to discuss homotopy groups, and thereby introducing new names for things I already knew. Thus, I give you, for a pointed topological space $$M$$

### The space of loops based at $$x_0$$: $$\Omega_{x_0}M$$

This is the quite normal set of all possible loops in $$M$$ as used when defining the fundamental group. My guess, however, is that the terminology here used also ties this familiar object into the loop spaces that people study.

### The Pontryagin product

This is the normal "First one loop, then the other" product of elements in $$\Omega_{x_0}M$$. I didn't know it carried that name though.

Using these new names, the fundamental group of a space $$M$$ is just $$(\Omega_{x_0}M/\sim,*)$$ with homotopy equivalence for $$\sim$$and the Pontryagin product for $$*$$.

These definitions are analogously extended to embeddings of $$S^n$$ instead of loops to form the higherdimensional homotopy groups. It's not clear to me, however (though probably most due to the time past since I read about homotopy groups closely) how to just extend the Pontryagin product to hyper-spheres. So you have two hyper-spheres that touch in the basepoint of the space. Hmmmm. If you remove that point, you end up with two punctured spheres (homeomorphic to disks) that stretch into the basepoint. So you glue the spheres together in the basepoint that you reinsert. So you get basically a very narrow hourglass. And this is homotopic to a sphere again... Yeah, I think I can visualize that.

A homeomorphism $$f$$ that takes $$M_1$$ to $$M_2$$ will also be a homeomorphism from $$M_1\setminus A$$ to $$M_2\setminus f(A)$$, since precisely those bits are removed which would mess with bijectivity, and continuity is preserved due to the set difference having the induced topology. This can be used together with the result (to be proven) that $$\pi_m(S^n)=0$$ for $$m Indeed, should there be such a homeomorphism, then that homeomorhpism would with this result give a homeomorphism [tex]\mathbb R^m\setminus 0$$ to $$\mathbb R^n\setminus f(0)$$. Both of these spaces are punctured euclidean spaces, and therefore homotopic equivalent to the corresponding spheres and thus have the same homotopy groups as the spheres do. But we know that homeomorphic spaces have equal homotopy groups, and that spheres of different dimensionality have different homotopy groups (since $$\pi_m(S^n)=0$$ but $$\pi_m(S^m)\neq 0$$) and thus the homeomorphism cannot exist.

## And the things I didn't know before

A diffeomorphism is a homeomorphism such that the homeomorphism and its inverse both are smooth. We need open sets in real vectorspaces as domain and codomain for this to work as defined. Diffeomorphisms come in classes: $$C^r$$ - which just denote how many times you can differentiate both the homeomorphism and its inverse.

A slightly weaker concept is that of local diffeomorphism - which is a map from an open set to an open set, as with diffeomorphisms, such that for each point there is an open neighbourhood such that the map is a homeomorphism of that neighbourhood onto its image.

Maps can be tested for local diffeomorphism status by calculating their Jacobian - the determinant of the matrix $$\left(\frac{\partial f^i}{\partial x^j}\right)$$. When the matrix degenerates, that is when the determinant is zero, at some point, then the map fails to be a diffeomorphism.

### Exercise / Example

The map $$f:\mathbb R\to\mathbb R$$ given by $$f(x)=x^3$$ is not a diffeomorphism. Indeed, the Jacobian is $$3x^2$$, which is $$0$$ for $$x=0$$.

#### Manifolds

A covering is a family of open sets such that the union of the sets contain the space.

A covering is locally finite if every point has an open neighbourhood which intersects only finitely many covering sets.

A space is paracompact if every covering has a locally finite subcovering. A space is compact if every covering has a finite subcovering.

A space if an $$n$$-dimensional topological manifold if it is Haussdorff and every point has a neighbourhood homeomorphic to an open subset of $$\mathbb R^n$$.

### Examples

The circle is a 1-manifold. So are many algebraic curves. However, not necessarily all 1-dimensional algebraic varieties - for instance the variety generated by the equation $$xy$$ fails, since every neighbourhood of $$(0,0)$$ by necessity has to be a cross-shape, which in no way is an open subset of any $$\mathbb R^n$$ and definitely not of $$\mathbb R$$.

A chart at a point $$x_0$$ of a topological manifold $$M$$ is a pair $$(U,\phi)$$ consisting of an open neighbourhood $$U$$ of $$x_0$$ and a homeomorphism $$\phi:U\to V$$ to an open set $$V\subset\mathbb R^n$$. The actual values in the image of a point are called the coordinates of the point. An atlas is a family of charts covering the space.

An example. Take the circle, embedded in $$\mathbb R^2$$ as the unit circle. I shall construct a few atlases on the circle and compare them.

For the first one, let the atlas be defined on four open (in the induced topology) subsets of the circle - namely $$x>0$$, $$x<0$$, $$y>0$$ and $$y<0$$ being the defining relations for each of the four. Each such subset is a halfcircle, excluding the endpoints. Furthermore, for the open subset defined by $$y>0$$ let a map to $$\mathbb R$$ be given as $$(x,y)\mapsto x$$.

First off: Is this a homeomorphism? It is a continuous map, quite clearly. Furthermore, it is a bijection onto the open subset $$(-1,1)$$ in $$\mathbb R$$. And the inverse, given by the function $$x\mapsto (x,\sqrt{1-x^2})$$ is continuous as well. Thus it is, indeed, a homeomorphism.

To get the other four charts, change x for y, and twiddle the sign of the square root in the inverse to choose which half-circle you want to get to. Everything checks out, mutatis mutandi.

For the second one, take the same partitions, but now use the positive radian distance to the x-axis instead. So the half-circle given by $$y>0$$ turns out to be mapped homeomorphically to $$(0,\pi)$$, $$x<0$$ maps to $$(\pi/2,3\pi/2)$$, $$y<0$$ maps to $$(\pi,2\pi)$$ and x>0 maps to $$(3\pi/2,5\pi/2)$$.

For a pair of charts that overlap, every point in the intersection has two sets of coordinates - one from each chart. Say the charts are $$(U_\alpha,\phi_\alpha)$$ and $$(U_\beta,\phi_\beta)$$. Then one set of coordinates are the coordinates of $$\phi_\alpha(x)=\{x_\alpha^i\}_{1\leq i\leq n}$$ and the other set of coordinates are the coordinates of $$\phi_\beta(x)=\{x_\beta^j\}_{1\leq j\leq n}$$. Thus for every pair of overlapping charts, we get a family of $$n$$ functions $$f^i_{\alpha\beta}:\mathbb R^n\to\mathbb R$$, each taking the point $$\phi_\alpha(x)$$ to the $$i$$:th coordinate in $$\phi_\beta(x)$$.

An atlas is said to be a smooth atlas if the transformation maps $$f_{\alpha\beta}$$ for each pair of overlapping charts is smooth.

For our examples, the transformation maps are as follows:

For the projection maps (the first example), the overlaps are in each quadrant. I'll work through the first quadrant, leaving the rest to be done equivalently. The point $$(x,y)$$ is mapped to $$x$$ and $$y$$ respectively in the two charts. So if we go from $$(0,1)$$ to $$(0,1)$$ from the $$x$$-axis up and then over to the $$y$$-axis, we would end up with the following map:
$$x\mapsto(x,\sqrt(1-x^2))\mapsto\sqrt(1-x^2)$$
This is indeed a smooth map, and all other maps are on the same form; so we have a smooth atlas.
For the radian distance atlas, we again take this quadrant as a comparison points, but for a different reason. In all other quadrants, the conversion map is simply the identity, but here something nontrivial occurs.
We start with the quadrant as a part of the $$y>0$$ chart and want to convert this to a point in the $$x>0$$ chart. So we have a point $$(x,y)$$ on the circle. This point has radian distance $$\varphi$$ from the $$x$$-axis. So our map is $$(0,\pi/2)\to(2\pi,5\pi/2)$$ by
$$\varphi\mapsto(x,y)\mapsto\varphi+2\pi$$
due to the way the $$x>0$$ chart is built.

Two smooth atlases are equivalent if their union is a smooth atlas. In this entire treatment, smooth can be replaced with analytic, class $$C^r$$ et.c. to yield a theory for such atlases.

So. Are my two atlases equivalent? The union is the atlas with charts all defined on four half circles, but with different transfers. Within each atlas, we can step back and forth smoothly between charts. So if I, again, treat one transfer from one halfcircle with the projection chart to the same halfcircle with the radial distance chart, that transfer can be extended analogously to the rest of the charts. For non-identical domains, we can always first step over using identical domains and then use the atlas-internal transfer to get where we want to go.
So, we take, again $$y>0$$ as our workspace. We have to charts, one which sends it to $$(-1,1)$$ and one which sends it to $$(0,\pi)$$. So we start with the point $$\varphi$$ in $$(0,\pi)$$. This goes to $$(\cos(\varphi),\sin(\varphi))$$, and then is sent to $$\cos(\varphi)$$. So that's smooth. For the other possible order, we go from $$(-1,1)$$ to $$(0,\pi)$$ sending $$x$$ to $$\arccos(x)$$. So everything checks out fine, and my atlases are equivalent.

Now. Is this an equivalence class? Reflexivity and commutativity hold straight off. The remaining question is that if the atlas $$\alpha$$ is equivalent to the atlas $$\beta$$, and $$\beta$$ with $$\gamma$$, are then $$\alpha$$ and $$\gamma$$ equivalent? For this to hold, $$\alpha$$ and $$\gamma$$ need to be compatible. But the transition from $$\alpha$$ to $$\gamma$$ could be made via the transfer functions from $$\beta$$ while ignoring $$\beta$$ as an atlas. So since compositions keep smoothness intact, we're home free.

Thus, we can form equivalence classes of atlases. A manifold with such an equivalence class is said to have a smooth (real analytic, class C^r) structure.

This concludes sections up to 1.8 of the notes. More to come later.