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2 Added a commutative algebra proof

I have a feeling that almost every field of pure mathematics has its "own" way to see that surface groups $\pi_g$ (of genus $g\ge 2$) are not free. The two arguments below are, of course, more complicated than the ones which are based on group theory, geometry, topology or homological algebra, but the idea is to see connection to other fields of mathematics. Here are two examples:

1. Algebraic geometry (combination of abelian and nonabelian Hodge theory): If $\pi_g\cong F_r$, free group of rank $r$, then, by looking at the 1-st Betti numbers and using the usual Hodge theory we see that $r$ has to be even. On the other hand, by Narasimhan-Seshadri theorem, the moduli space ${\mathcal M}_g$ of semistable rank 2 holomorphic bundles (with fixed determinant) on genus $g$ surface is analytically isomorphic to $Hom(\pi_g, SU(2))/SU(2)=Hom(F_r, SU(2))/SU(2)$, which would have odd (real) dimension $3r-3$. Contradiction, since (being a complex-projective variety) ${\mathcal M}_g$ is even-dimensional. (This is, of course, an argument similar to, but more complicated, than Mohan's.)

2. Geometric analysis: Suppose that $\pi_g\cong F_r$. Realize this isomorphism $\rho$ by a harmonic map $h$ from $S_g$ (genus $g$ compact Riemann surface) to a metric graph $\Gamma_r$ (the rose with $r$ leaves and unit edges). Preimages of generic points in $\Gamma_r$ under $h$ will be compact 1-dimensional submanifolds. By the maximum principle (look at the lifted harmonic map from the universal cover of $S_g$ to the tree), components of these submanifolds cannot bound disks in $S_g$, hence, they are not nul-homotopic. Hence, $\rho$ cannot be injective.

So, are there proofs (even difficult ones) which make essential use of other fields of mathematics, e.g.:

a. Number theory (algebraic or analytic number theory, or arithmetic algebraic geometry)?

b. Measure theory? (This might be difficult since $\pi_g$ and $F_r$ are "measure-equivalent".)

c. Probability? (An argument using random walk on graphs maybe?)

d. Dynamical systems/ergodic theory?

e. Functional analysis? (Maybe infinite-dimensional unitary representations of $\pi_g$ and $F_r$?)

f. Logic?

g. Commutative algebra?

Update: Here is a proof by the commutative algebra. Consider the affine schemes $S=Hom(\pi_g, GL(2))$ and $S'=Hom(F_r, GL(2))\cong GL(2)^r$. The latter, being an open subscheme of the affine space, has the same dimension of Zariski tangent space at every point. Consider $S$, let $R$ be its coordinate ring and $m_1, m_2\subset R$ be the ideals corresponding to the points $\rho_1$ (the trivial representation) and $\rho_2$, the representation which sends all but one standard generators to $1$ and the remaining generator to any noncentral matrix in $GL(2)$. Then (by a reasonably simple computation) $d_1=dim(R/m_1)=8g$ while $d_2=dim(R/m_2)=8g-2$. Hence, $d_1>d_2$ and dimensions of Zariski tangent spaces to $S$ at $\rho_1, \rho_2$ are different. In particular, the schemes $S, S'$ (equivalently, their rings) cannot be isomorphic and $\pi_g$ cannot be isomorphic to $F_r$.

1

I have a feeling that almost every field of pure mathematics has its "own" way to see that surface groups $\pi_g$ (of genus $g\ge 2$) are not free. The two arguments below are, of course, more complicated than the ones which are based on group theory, geometry, topology or homological algebra, but the idea is to see connection to other fields of mathematics. Here are two examples:

1. Algebraic geometry (combination of abelian and nonabelian Hodge theory): If $\pi_g\cong F_r$, free group of rank $r$, then, by looking at the 1-st Betti numbers and using the usual Hodge theory we see that $r$ has to be even. On the other hand, by Narasimhan-Seshadri theorem, the moduli space ${\mathcal M}_g$ of semistable rank 2 holomorphic bundles (with fixed determinant) on genus $g$ surface is analytically isomorphic to $Hom(\pi_g, SU(2))/SU(2)=Hom(F_r, SU(2))/SU(2)$, which would have odd (real) dimension $3r-3$. Contradiction, since (being a complex-projective variety) ${\mathcal M}_g$ is even-dimensional. (This is, of course, an argument similar to, but more complicated, than Mohan's.)

2. Geometric analysis: Suppose that $\pi_g\cong F_r$. Realize this isomorphism $\rho$ by a harmonic map $h$ from $S_g$ (genus $g$ compact Riemann surface) to a metric graph $\Gamma_r$ (the rose with $r$ leaves and unit edges). Preimages of generic points in $\Gamma_r$ under $h$ will be compact 1-dimensional submanifolds. By the maximum principle (look at the lifted harmonic map from the universal cover of $S_g$ to the tree), components of these submanifolds cannot bound disks in $S_g$, hence, they are not nul-homotopic. Hence, $\rho$ cannot be injective.

So, are there proofs (even difficult ones) which make essential use of other fields of mathematics, e.g.:

a. Number theory (algebraic or analytic number theory, or arithmetic algebraic geometry)?

b. Measure theory? (This might be difficult since $\pi_g$ and $F_r$ are "measure-equivalent".)

c. Probability? (An argument using random walk on graphs maybe?)

d. Dynamical systems/ergodic theory?

e. Functional analysis? (Maybe infinite-dimensional unitary representations of $\pi_g$ and $F_r$?)

f. Logic?

g. Commutative algebra?