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Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

[Added later]: My paper on WZW models and $H^4(BG,\mathbb Z)$ has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$.

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which it follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

Recently, prompted by considerations in conformal field theory, I was lead to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

[Added later]: My paper on WZW models and $H^4(BG,\mathbb Z)$ has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$.

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which it follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which it follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

[Added later]: My paper on WZW models and $H^4(BG,\mathbb Z)$ has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$.

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which it follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free, seem to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which is follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.

Recently, prompted by considerations in conformal field theory, I was let to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free.

By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result.

However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following:

Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free?

For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation].
Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument).

Now, here comes the computation:
$H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$
from which it follows that for any finite abelian group $A$
$H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$
from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$
$H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$

The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED

PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is.