We have some space XX with a basepoint x0x_0, and define π1(X,x0):=π((S1,0),(X,x0))=cone of ΩX\pi_1(X, x_0) := \pi((S^1,0),(X,x_0)) = \text{cone of }\Omega X. path conected.

You might think a group has one operation but a group actually has three operations, one binary which is the composition, one unary which is the inverse, and one nullary which is the identity. Constant loop will play the roll of the identity element, have reverses of loops be called invereses (γ−1(t)=γ(1−t)\gamma^{-1}(t) = \gamma(1-t)), and multiplication is defined by composition of loops.

If we have a loop about x1x_1 and a path from x0x_0 to x1x_1, then we can get a loop about x0x_0 by conjugation in this group of loops. By the way, the loop of groups here is what is called the fundamental group. What we can say is that:

π1(XCW,x0)=π1(sk2XCW,sk0)\pi_1(X^{CW},x_0) = \pi_1(\text{sk}_2X^{CW},\text{sk}_0)
and by definition, if XX is simply connected, then π1(X,x0)={e}\pi_1(X,x_0) = \{e\}. In particular, for n≥2n \ge 2, π1(Sn≥2)={e}\pi_1(S^{n\ge 2}) = \{e\}.

Remark. Mentioned sometihng about needing to check that you can make one point free? For sphere stuff, but said that we have already done that.

Coverings (and Fundamental Groups)

Example.

Definition. A covering is just a map pp from some other space TT to XX (usually written vertically rather than horizontally). Gneerally we think XX is path connected.

Beware that covering is not the same of just a homeomorphism or something I might have to clarify later what the warning was.

Then we overed the Homotopy Lifting Lmema.

So if you break the rectangle into grids, you can find a erfined enough partition, apparently some similar proof to Heine-Borel Theorem, way to use compactness of the cube.

What we can do now is that ?

To prove this, we first take a Lemma where Z=I0Z = I^0, I1I^1, or InI^n in general. This Lemma is proved via the diagram I drew in Krita.

A polyhedron is something like {x:Ax≤b}\{x : Ax \le b\}. AA is a m×nm \times n matrix.

We can get a map by just tracing any point in the cylinder back to the base of the cylinder, and then mapping that back via the lemma. So now the question is more verifying that it is continuous.

Something like “if we define the cylinders small enoguh, they each land in a defining neighborhood in XX”.

Something analogous in multivariable calculus where given aa and bb we want to find some ff such that

df=a  dx+b  dydf = a \;dx + b\;dy
and to get this we can draw a square, integrate first over yy and then integrate over xx.

I think he’s discussing the theorem in 6.6

Moreover: p−1(x0)≃p∗π1(T1?)\forwardslashπ1(X,x0)p^{-1}(x_0) \simeq p_*\pi_1(T_1?) \forwardslash \pi_1(X,x_0) right cosets.

Now we do the map-lifting theorem. There is a map defined in the only possible way and we need to consider continuity.

Remark. Intuitively, I feel like two disjoint unions of opens should be homeomorphically equivalent if and only if they have the same number of path connected components.

Finally getting to criterion for equivalence of coverings.

If we have two coverings one from TT and one from T′T', then we can take map-lifting theorem to get two homeomorphisms between TT and T′T'.

If XX is semilocally simply connected, then there exists a unique universal covering from a simply connected space. We can then do a construction of the universal covering.

Next Lecture…

I think E(X,x0)E(X,x_0) denotes the universal covering quotiented by something.

This discussion is supposedly related to the third problem on the homework.

What is the inverse image of this thing in the loop space? It consists of all paths which end in VV and whichif I connected the endpoint with that point by a path inside UU, the composite path is homotopic to γ\gamma.

Compact open condition…

From this we get that the universal cover E(X,x0)/(γ∼γ~)E(X,x_0)/(\gamma \sim \tilde{\gamma}) is the universal cover of this space.

Now we define Deck Transformations.

π1(X,x0)\pi_1(X,x_0) acts freely and discretely on X~\tilde{X}, and X=X~/π1(X,x0)X = \tilde{X}/\pi_1(X,x_0). This is supposedly obvious now, but I don’t get it.

Consider π⊆π1(X,x0)\pi \subseteq \pi_1(X,x_0). We have this universal cover X~→X\tilde{X} \rightarrow X

Moreover, π\pi acts ferely and discretely on a simply connectd YY.

He drew some tree and said tree contractible, maybe you need some technology and stuff. What is clear is that any finite tree is contractible. Also related to some words. Every loop is compact so it lands in a finite part of it.

We then talked about abstract deck transformations, and mentioned something like p∗(γt0t1)∈N(π)/πp_*(\gamma_{t_0}^{t_1}) \in N(\pi)/\pi. There was a theorem, map lifting theorem, existence of map, image of fundamental group here lies in image of that group.

If π⊆π1(X,x0)\pi \subseteq \pi_1(X,x_0) is normal, then the corresponding covering T,t0→pX,x0T,t_0 \xrightarrow{p} X,x_0 is called regular.

ITS JOVER BRUH HES BRINGING GALOIS THEORY INTO THIS… my worst nightmare comes back 😭

Example (Related to Galois theory). H⊆Cx,a1,...,ann+1H \subseteq \mathbb{C}_{x,a_1,...,a_n}^{n+1}, consider the hypersurface {xn+a1xn−1+...+an=0}\{x^n+a_1x^{n-1}+ ... + a_n = 0\}. For example with x3+a2x+a3=0x^3 + a_2x + a_3 = 0, you can always translate the space so that the sum of the roots is zero (why we can assume a1=0a_1 = 0). Hypersuurface is mapped to discriminant Δn−1⊆\Cn\Delta^{n-1} \subseteq \C^n or something.

For each polynomial with nn roots there are nn roots so it is an nn-fold cover.

Consider the field of rationals C(a1,...,an)\mathbb{C}(a_1,...,a_n). There isa field extension by adjoining xx to this field, C(a1,...,an,x)\mathbb{C}(a_1,...,a_n, x). These are rational functions on this hypersurface HH. They form a field and its a field extension. We have a map from Cx1,...,xnn\mathbb{C}^n_{x_1,...,x_n} to Ca1,...,ann\mathbb{C}^n_{a_1,...,a_n} via (−1)kσk(x1,...,xn)(-1)^k \sigma_k(x_1,...,x_n), elementary symmetric polynomials or something.

If we quotient out by Δn−1\Delta^{n-1}, we get a regular covering or something, SnS_n or something. Or something.

C(a1,..,an)⊆C(a1,...,an,x)⊆C(x1,...,xn+1?)⊇Sn\mathbb{C}(a_1,..,a_n)\subseteq \mathbb{C}(a_1,...,a_n,x) \subseteq \mathbb{C}(x_1,...,x_{n+1?}) \supseteq S_n.

He drew many triangles on the board, along with an equation x1+x2+x3=0x_1 + x_2 +x_3 = 0. Apparently that’s important and we can see it on the real plane.

“Now let us try to figure out the fundamental group of the discriminant. The name for it is a braid group on nn strings.”

First of all you need to pick a basepoint on the complement of the discriminant. Then you need to take aloop around the discriminant, and each point is a polynomial, and the roots start moving somehow. You get some parallel planes and the roots somehow move and when they come back they come back to the same position.

He drew a picture with strings going from nn points above to nn points on the bottom crossing somehow and deciding which goes over and under, and we can slice it up so that each slice only has one crossing.

Apparently the relationship was between regular coverings of x1,...,xn−∪(xi=xj)x_1,...,x_n - \cup (x_i = x_j) over SnS_n are somehow analogous to normal extensions of C(x1,...,xn)\mathbb{C}(x_1,...,x_n) over C(a1,...,an)\mathbb{C}(a_1,...,a_n).