Are Same-Action Deck Transformations & Loops Equal for S1 x S1?

[mich0144]

I'm self studying some alg topology for next semester just working through chapter 0 and 1 of hatcher really. My question is: for any universal cover p of X there are two actions of pi_1(X, x0) on the fiber p^-1(x0) given by lifting loops at x0 and given by restricting deck transformations to the fiber. are they the same for S1 x S1? would pi_1(X, x0) being abelian help?

So an action G on X is (G,X) -> X. so here G would be pi_1(X, x0) and X would be some point in p^-1(x0) and this action gives another point in p^-1(x0). The deck transformation is defined to be a homeomorphism of the covering space with itself (where universal cover of S1 x S1 is R x R here). So I can see what both are doing geometrically but what does it for the actions to be the same really? how do i tackle this.

I'm also not sure about abelian, I think it means f*g and g*f have to be homotopic where f,g, are two loops in pi_1(X, x0) where is this useful usually.

 

[mich0144]

any idea I think in general the two actions are not the same not sure about the torus or the wedge of 2 circles, I'm using the basic universal cover R of S1 so p:(cos2pix,sin2pix)

 

[quasar987]

The actions are the same for if \tilde{X} is the universal cover of X, then G(\tilde{X}) = pi_1(X). (The covering group is isomorphic to pi_1). And this isomorphism is the map that takes a class of loops [f] in X based at x_0 to the unique deck transformation \varphi_f which takes \tilde{x_0} to F(1), where F is the unique lift of f with F(0)=\tilde{x_0}. (See the proof of Prop 1.39)

So action of pi_(X,x_0) on the fiber by [f] is the same as action of G(\tilde{X}) on the fiber by \varphi_f.

 

[mich0144]

So you're saying the two actions are always the same if we take p to be the universal map, what characteristic of universal map are you using to conclude this and what's a counterex. for a nonuniversal map.

the question (27 in hatcher) specifically asks about s1 V s1 and s1 x s1 and then ask if they are always true for abelian fundamental groups. So the questions makes me a little suspicious about this resting on universality of covering maps alone since s1 x s1 is abelian and s1 V s1 isn't.

 

[quasar987]

The beginning of the question is this:

"For any universal cover p of X there are two actions of pi_1(X, x) on the fiber p^-1(x) given by lifting loops at x and given by restricting deck transformations to the fiber."

So, what is he talking about here? The first action is pretty clear: fix a lift \tilde{x} of x, then consider the action mapping a loop class [f] in pi_1(X,x) to F(1), where F is the unique lift of f such that F(0)=\tilde{x}.

The second action is a little more cryptic: "restricting deck transformations to the fiber"?! What's pi_1(X,x) has to do with that?
Well, the answer is provided by Prop 1.39 which says that G(\tilde{X})\approx \pi_1(X,x) via the map \varphi:\pi_1(X,x)\rightarrow G(\tilde{X}) of the proof which sends [f] as above to the unique deck transformation that maps \tilde{x} to F(0).
So the second action would be this: \pi_1(X,x) \times p^{-1}(x)\rightarrow p^{-1}(x) : ([f],\tilde{x})\mapsto \varphi([f])\cdot\tilde{x}=F(1), where F is the unique lift of F such that F(0)=\tilde{x}.

So, yeah the two actions are trivially the same and the question in Hatcher is stupid, unless there is a different but equally natural way to interpret Hatcher's cryptic "the action of pi_1(X, x) on the fiber p^-1(x) given by restricting deck transformations to the fiber". Can you think of one? (Not me.)

 

[mich0144]

Thanks in your definition as well as in hatcher they show the deck transformation to be a left group action and lift to be a right this is what's confusing me, why is this so. I thought it was purely notational can't you write either in the other way.

 

[quasar987]

Mmmh, yes it doesn't matter either left or right action, it's formally the same thing.

But surely there's something big and obvious we're missing here. :(

 

[mich0144]

Yea I think they are infact the same group like you said demonstrated by the isomorphism but the actions act differently, I'm not sure but maybe converting right to left action has something to do with abelianization so. I need to review my algebra this problem can't be so trivial.

 

[quasar987]

Please tell me if you find something.