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I apologize if the subject doesn't accurately describe my question.

Let $F_2$ denote the free group on two generators.

Suppose you have some group homomorphism $A : \mathbb{Z}^2\rightarrow\mathbb{Z}^2$. There are many ways to lift $A$ to a map $\alpha : F_2\rightarrow F_2$ (ie, the square diagram with vertical maps being abelianizations, commutes).

Firstly, am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$?

Now let $\alpha : F_2\rightarrow F_2$ denote a particular lift of the map on $\mathbb{Z}^2$ sending $(x,y)\mapsto(mx,my)$ for some positive integer $m$.

Let $E$ be a complex elliptic curve (ie, a complex torus). Let $[m] : E\rightarrow E$ denote the multiplication by $m$ map. Let $E^*$ denote the same curve, with the identity point removed. We identify $F_2$ with $\pi_1(E^*)$, and $\mathbb{Z}^2$ with $\pi_1(E)$, with suitable base points. The inclusion maps $E^*\hookrightarrow E$ on the level of fundamental groups become exactly abelianization maps $F_2\rightarrow \mathbb{Z}^2$. The map $[m]$ gives some map $\mathbb{Z}^2\rightarrow\mathbb{Z}^2$, so we're in a similar situation to what I described earlier.

Since the universal cover of $E,E^*$ are $\mathbb{C},\mathcal{H}$(the upper half plane) respectively, both are Eilenberg Maclane spaces. By a theorem described in Hatcher, if $Y$ is an Eilenberg-Maclane space, and $X$ a CW complex, then for homomorphism $\pi_1(X,x_0)\rightarrow\pi_1(Y,y_0)$, there is a map $X\rightarrow Y$ sending $x_0$ to $y_0$ which induces the given map on fundamental groups. Furthermore, this map is unique up to homotopy fixing $x_0$.

My questions are as follows:

  • Is the theorem true for $X = E^*$? (ie, is it true that $E^*$ can be realized as a CW-complex or perhaps the theorem can be generalized to include the case $X = E^*$?)
  • Assuming the answer to (1) is yes, there should be a map $E^*\rightarrow E^*$ inducing $\alpha$ on fundamental groups. How can we best describe this map? This map certainly cannot be a restriction/lift of $[m]$, since every $m$-torsion point would want to map to the identity of $E$, which is not in $E^*$. On the other hand, this map is "unique", which generally says it should somehow be close to a restriction/lift of $[m]$ to $E^*$...

Any help would be appreciated.

Thanks,

Will

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Am I right in saying that any two lifts of $A$ to an endomorphism of $F_2$ differ by some inner automorphism of $F_2$?

No, using a free basis $F_2 = \langle a,b \rangle$, the map given by $f(a)=a$, $f(b) = a b^3 a^3 b^{-1} a^{-4} b^{-1}$ is not an inner automorphism but it induces the identity on $\mathbb{Z}^2$, as does the identity automorphism $g(a)=a$, $g(b)=b$.

Is the theorem true for $X = E^*$? (ie, is it true that $E^*$ can be realized as a CW-complex or perhaps the theorem can be generalized to include the case $X = E^*$?)

The punctured torus $E^*$ deformation retracts to a 1-dimensional complex $R_2$, the "rank 2 rose" consisting of two loops touching at a point, namely the union of an $a$-curve and a $b$-curve in $E^*$. So yes, $E^*$ is an Eilenberg-Maclane space, because it is homotopy equivalent to the 1-complex $R_2$.

Assuming the answer to (1) is yes, there should be a map $E^*\rightarrow E^*$ inducing $\alpha$ on fundamental groups. How can we best describe this map?

There are many such maps, but one has no expectation that they are anything other than yucky. The lifted endomorphisms of the map $[m]$ are many and varied. For instance $f(a)=a^m$, $f(b)=b^m$ is one, but also $g(a)=a^m$, $g(b) = b a^{42} b a^{71} b^{m-2} a^{-113}$. If you represent each of these formulas as a topological map on $R_2$ then you get a map on $E^*$ as a composition $$E^* \to R_2 \to R_2 \to E^* $$ where the first map is the deformation retraction, the second is the formula $f$ or $g$, and the third is the inclusion. Yuck.

On the other hand, this map is "unique", which generally says it should somehow be close to a restriction/lift of $[m]$ to $E^*$…

Quite the opposite. Whatever this map might be, as I have tried to demonstrate it is very far from "unique".

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  • $\begingroup$ I don't think your definition $f(a) = a, f(b) = ab^2a^3b^{-1}a^{-2}b^{-1}$ induces the identity on $\mathbb{Z}^2$, but I think defining $f(a) = bab^{-1}$ and $f(b) = aba^{-1}$ should induce the identity on $\mathbb{Z}^2$, despite not being an automorphism, I think... Also, the map Aut($F_2) \rightarrow\text{GL}(2,\mathbb{Z})$ does actually have the inner automorphisms as kernel: en.wikipedia.org/wiki/Out(Fn) $\endgroup$ – oxeimon Apr 14 '14 at 3:17
  • $\begingroup$ You are right about my faulty addition, and about my mistake regarding the kernel of $Aut(F_2) \to GL(2,\mathbb{Z})$. I have corrected each of them. $\endgroup$ – Lee Mosher Apr 14 '14 at 20:16

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