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\title{Borel's proof of Beilinson's Conjecture for number fields}

\author{Akshay Venkatesh \footnote{Notes by Tony Feng}}

\date{May 4, 2016}

\begin{document}

\maketitle

As we have discussed, Beilinson conjectureed that an $L$-value is a regulator on motivic cohomology. The only case in which anything is known is for $\zeta_F$ for $F$ an imaginary quadratic field. By this we mean that Borel gave a relation between the values of $\zeta_F$ at integers and regulators on $K_* \Cal{O}_F$. (Some further work is required to identify this with Beilinson's conjecture, i.e. that the Borel regulator really is the Beilinson regulator in this case.) 

\section{Hopf algebras}
\subsection{Cohomology of compact Lie groups}
Recall that for the unitary group $U_n$, the cohomology ring is an exterior algebra on odd generators:
\[
H^*(U_n) = \CC[e_1, e_3, e_5, \ldots, e_{2n-1}].
\]
We will need this later so we review why this is true. 

\begin{example}
Let $A = \Q[G]$ for a finite group $G$. This is a Hopf algebra. Let's not fuss about the general definition of a Hopf algebra; suffice it to say that it reflect extra structure, which we can see because we can tensor and dualize representations of $A$. The ability to tensor is explained by a \emph{coproduct} 
\[
\Delta \colon A \rightarrow A \otimes A
\]
sending $g \mapsto g \otimes g$. The ability to dualize is explained by a map $A \rightarrow A$ sending $g \mapsto g^{-1}$.

The dual algebra $A^{\vee} = \text{Functions}(G, \Q)$ is also a Hopf algebra because the dual of the coproduct is a product structure, and the dual of a product is a coproduct structure. 

Note that $A$ is a non-commutative algebra, but has commutative coproduct. On the other side, the product is commutative and the coproduct is non-commutative. 
\end{example}

If $G$ is a compact Lie group, then $H^*(G, \CC)$ is a Hopf algebra. The coproduct structure comes from the multiplication map 
\[
m \colon G \times G \rightarrow G.
\]
Also $H_*(G, \CC)$ is a Hopf algebra, with product coming from $m \colon G \times G \rightarrow G$ and coproduct induced by 
\[
\Delta \colon G \rightarrow G \times G
\]
sending $g \mapsto (g,g)$. 

\begin{prop}
If $G$ is compact connected then $H_*(G, \CC)$ is commutative and $H^*(G, \CC)$ is co-commutative. 
\end{prop}

\begin{proof}
Fix a Riemannian metric on $G$ which is left and right invariant. By de Rham theory, 
\[
H^*(G, \CC) \cong \{ \text{harmonic forms on $G$}\}. 
\]
We claim that a harmonic form on a compact group must automatically be invariant; denote the invariant forms by $\Inv(G)$. Why? Translation by $G$ can only act trivially on cohomology, but there is a unique harmonic representative for each de Rham cohomology class. This shows that harmonic forms are contained in $\Inv(G)$.

We next claim that in fact this is an equality: $\{ \text{harmonic forms on $G$}\} = \Inv(G)$. How can we see this? We want to show that if $\omega \in \Inv(G)$ then $d\omega = 0$. Well, an invariant form is determined by its value at the identity, so it suffices to study the Lie algebra. The effect of inversion $i (x) =  x^{-1}$ on the group acts by $-1$ on the Lie algebra. Therefore, it acts by $(-1)^{p}$ on forms of degree $p$. Since $d$ is equivariant for it, we have 
\[
(-1)^{p+1} d \omega =  i^* d \omega = d i^* \omega = (-1)^p \omega
\]
which shows that $d \omega = 0$. Applying the same argument to $* \omega$ shows that $\omega$ is both closed and coclosed, hence harmonic. 

Now consider the commutative diagram
\[
\xymatrix{
G \times G \ar[r]^{xy}  \ar[d]_{i \times i} & G \ar[d]^i \\
G \times G \ar[r]^{yx}  & G
}
\]
Since we have identified the cohomology with invariant forms, we know that $i$ acts as $(-1)^{\deg}$ on cohomology, this shows the graded commutativity of the Hopf algebra structure.
\end{proof}

\begin{thm}[Milnor, Moore] Suppose $A = \bigoplus_{i \geq 0} A_i$ is a graded commutative, graded cocommutative Hopf algebra over a field $k$ in characteristic $0$ (so $A_0=k$). Then $A$ is free on the subspace $\Prim(A)$, which is $\{ a \in A \colon \Delta a = a \otimes 1 + 1 \otimes a\}$. 
\end{thm}

In particular, if $A$ is finite-dimensional then there can be no generators in even degree, so $A$ must be a free exterior algebra on elements in odd degree. This explains why the cohomology of a compact Lie group must be free on odd generators.\\

\begin{example}
If $A$ were in even degree then $\Spec A$ would be a commutative algebraic group $G$. Then $G$ is a product of $\G_a$'s and $\G_m$'s but we claim that in fact $G \cong \G_a^N$. This is because the graded structure on $A$ implies that there is a $\G_m$-action on $G$, and there cannot be a $\G_m$-action on the toral part. So $G \cong \G_a^N$. Therefore, $\Prim(A)$ can be identified with $\{ \chi \colon G \rightarrow \G_a\}$. 
\end{example}

\subsection{Invariant forms}

Let's go back to $\Inv(U_n)$. These are completely determined by what happens at the identity, and they have to be conjugacy-invariant there. In other words, they are alternating $j$-forms $\bigwedge^j \mf{u}_n \rightarrow \CC$ which are invariant under conjugation by $U_n$. This is the same as alternating forms $\bigwedge^j M_n(\CC) \rightarrow \CC$ which are invariant by $\GL_n \CC$. 

Here is one such form: $X \mapsto \Tr(X)$. More generally, for any $k$ we can consider $\Tr(X_1, \ldots, X_k)$ - this is a conjugacy-invariant form, but it is not alternating. To amend that, we anti-symmetrize it; but that is automatically $0$ if $k$ is even. So the candidate forms are $v_k(X_1, \ldots, X_k) =$ anti-symmetrization of $\Tr(X_1, \ldots, X_k)$. It turns out that these are generators for $\Inv(U_n)$. 

Why are they primitive? Consider for instance $v_3 = \Tr(XYZ) - \Tr(XZY)$. Pull back this form via the multiplication map
\[
U_n \times U_n \xrightarrow{x,y \mapsto xy} U_n.
\]
Unfortunately it is not clear that the pullback is still an invariant form. One would like to argue this by saying that the diagram 
\[
\xymatrix{
U_n \times U_n \ar[d]_{t_x \times t_y} \ar[r]^m & U_n \ar[d]_{t_{xy}} \\
U_n \times U_n \ar[r] & U_n }
\]
commutes, but actually it doesn't (because of the non-commutativity of $U_n$). 

To fix this, consider embedding into a larger group
\[
\xymatrix{
U_n \times U_n \ar[d]_{I \times I} \ar[r] & U_n \ar[d]^I \\
U_{2n} \times U_{2n} \ar[r] & U_{2n} 
}
\]
with the embedding $I \colon U_n \hookrightarrow U_{2n}$ being 
\[
g \mapsto \begin{pmatrix} g \\ & 1 \end{pmatrix}.
\]
This is conjugate to the embedding 
\[
J \colon g \mapsto \begin{pmatrix} 1  \\ & g \end{pmatrix}.
\]
So the diagram above is homotopic to 
\[
\xymatrix{
U_n \times U_n \ar[d]_{I \times J} \ar[r] & U_n \ar[d]^I \\
U_{2n} \times U_{2n} \ar[r] & U_{2n} 
}
\]
Since the forms $v_n$ on $U_n$ are the restriction of the analogous $U_{2n}$-invariantforms on $U_{2n}$, this shows that their pullbacks to $U_n \times U_n$ are in fact invariant. 

\section{$K$-theory of number fields}

Let $R$ be any commutative ring. Consider 
\[
\GL_{\infty}(R) := \varinjlim \GL_N(R). 
\]
Then $H_*(\GL_{\infty} R, \Q)$ is a graded commutative cocommutative Hopf algebra. The point is that it has a product, via the construction 
\[
\GL_n \times \GL_n \rightarrow \GL_{2n}. 
\]


There is a map
\[
K_i R \rightarrow \Prim(H_i \GL_{\infty}(R), \Q).
\]
For $i>0$, this is an isomorphism after tensoring with $\Q$, at least when $R = \Cal{O}_F$. So for our purposes, we can think of $K$-theory as this piece of group homology. 

\begin{remark}
The (group) homology of 
\[
\ldots \GL_N \Cal{O}_F \hookrightarrow \GL_{N+1} \Cal{O}_F \hookrightarrow 
\]
stabilizes. Quillen proved this to show that $K$-theory stabilizes, hence the $K$-theory of rings of integers is finitely generated. 
\end{remark}

Let $F$ be an imaginary quadratic field and $\Cal{O}$ is the ring of integers. Borel showed that
\[
(K_i \Cal{O})_{\Q} = \begin{cases} \Q & i \text{ odd } \geq 3 \\ 0 & \text{other } i>0 \end{cases}
\]
Also, there is a natural map 
\[
(K_i \Cal{O})_{\Q} \rightarrow \R
\]
called the \emph{Borel regulator}. (So called because $K_1$ is the unit group, and for $i=1$ this specializes to the classical regulator.) Borel prove that the image is $\sim \Q \cdot \zeta_F(\frac{i+1}{2})$. These zeta values arise as the volumes of $\SL_n \Cal{O}_F \backslash \SL_n \CC$. 

How do we compute $H^*(\SL_N \Cal{O}, \CC)$? We have an action of $\SL_N \Cal{O}$ on $\SL_n \CC/\SU(n)$, which is a contractible symmetric space $S_n$ of unitary forms. Then 
\[
H^*(\SL_n \Cal{O}, \CC) = H^*(S_n/\SL_n \Cal{O}, \CC).
\]
This $S_n$ comes with an invariant metric, and for it $X_n := S_n/\SL_n \Cal{O}$ has finite volume (but it not compact). (This was the reason for switching from $\GL_n$ to $\SL_n$.) Consider $\Inv^*(S_n)$, the space of invariant forms on $S_n$. This sits inside differential forms on $X_n$. 

Borel showed that the inclusion of $\Inv^j$ (meaning left $\SL_n \CC$-invariant $j$-forms on $\SL_n \CC/\SU_n$)  into differential forms on $X$ induces an isomorphism in $H^*$ when $j \leq n/4$. As before, $\Inv^j \subset \Harm^j$. The harmonic forms sit inside differential forms on $X_n$. We'd like to use Hodge theory to go backwards, but here we can't because our $X$ isn't compact. What Borel does is to consider an intermediate space of forms of ``moderate growth''. More precisely, replace $X_n$ by some truncation, and impose growth conditions on forms so that they lie $L^2$. 

The inclusion $\Inv^j \subset \Harm^j$ is only an equality in low degree. Since $S_n = \SL_n \CC/\SU_n$, an invariant $j$-form in $\Inv^j(S_n)$ is a function $\bigwedge^j T_e \rightarrow \CC$ which is invariant by $\SU_n$, where $T_e = \mf{sl}_n \CC / \mf{su}_n$. Since $\mf{su}_n$ is a real structure of $\mf{sl}_n \CC$, we can identify this with $i \mf{su}_n$. Thus we are looking at $\{ \bigwedge^j (i \mf{su}_n) \rightarrow \CC\}^{\SU_n}$, and since trace is no longer informative this is $\CC[\nu_3, \nu_5, \ldots, \nu_{2n-1}]$. 

In conclusion, we have found that in low degree $H^*(\SL_n \Cal{O}, \CC)$ is free exterior in degrees $3,5, \ldots$. Therefore $H^*(\SL_{\infty} \Cal{O}, \CC)$ is $\CC[e_3, e_5, e_7, \ldots]$. So $\Prim H^j (\SL_{\infty} \Cal{O}, \CC)= \CC$ for $j=3,5, \ldots$. 

We emphasize that the generators here are naturally indexed by the corresponding cohomology classes of the unitary group. 1

The same discussion applies for homology. 
 
 \section{The Borel Regulator}
Now we come to the second part, which is the regulator map 
\[
(K_i \Cal{O})_{\Q} \rightarrow \R.
\]
We have a map 
\[
\Prim H_j (\SL_{\infty} \Cal{O}_F , \Q) \xrightarrow{ \int \nu_j } \CC
\]
where $\nu_j$ is regarded as a cohomology class via 
\[
\nu_j \rightarrow \Inv^j \rightarrow H^j(\SL_n \Cal{O}, \CC).
\]
What is the image? We need to construct some explicit classes on the LHS. Where can we get explicit cycles? If $G$ is a group over $F$ embedded in $\SL_N$, then we get a map from the locally symmetric space for $G$ to the locally symmetric space for $\SL_n =: X_N$. We can then push forward the fundamental class for the locally symmetric space attached to $G$. 

\begin{example} Let $D$ be a division algebra of rank $4$ over $F$. Fixing an order $\Cal{O}_D$, consider the norm-1 units $\Cal{O}_D^{(1)}$. We can embed this in $\SL_n \Cal{O}_F$ by its action on $D$. This gives a map of symmetric spaces, 
\[
\pi \colon \Cal{O}_D^{(1)} \backslash \SL_2 \CC / \SU_2 \rightarrow X_N.
\]
The left hand side is 3-dimensional; let the fundamental class be $\mu$. Then the image of the Borel regulator for $i=3$ contains 
\[
\langle \pi_* \mu, \nu_3 \rangle = \int_{ \Cal{O}_D^{(1)} \backslash \SL_2 \CC / \SU_2} \pi^* \nu_3.
\]
This is basically $\zeta_F(2)$, up to factors of $\pi$. 

Now suppose $D$ is a divison algebra of rank $9$ over $F$. Then $\dim \Cal{O}_D^{(1)} \backslash \SL_2 \CC / \SU_2 = 8=3+5$. The image of the Borel regulator for $i=9$ contains 
\[
\langle \pi_* \mu, \nu_3 \wedge \nu_5 \rangle   =\zeta_F(2) \zeta_F(3).
\]
Let $e_3, e_5$ be generators for $\Prim H_i(\SL_{\infty})$. Then $\pi_* \mu = e_3 e_5 + \ldots$ with the extra stuff being higher-degree. (This happens in every degree by primitivity). So 
\[
\langle \pi_* \mu , \nu_3 \wedge \nu_5 \rangle = \langle e_3, \nu_3 \rangle \langle e_5, \nu_5 \rangle.
\]
We computed the first term to be $\zeta_F(2)$, and the product was computed to be $\zeta_F(2) \zeta_F(3)$, so we can deduce that $\langle e_5, \nu_5 \rangle \sim \zeta_F(3)$. 
\end{example}


\end{document}