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It is known that $H^k(SL(n,\mathbb{Z}))$ is independent of $n$ for $n \gg k$, so we can define a stable cohomology ring $$V = \text{lim}_{n \rightarrow \infty} H^{\ast}(SL(n,\mathbb{Z});\mathbb{R}).$$ Using analytic tools, Borel proved that $V$ is an exterior algebra generated in degrees $3, 5, 7, \ldots$. Is there proof of this using more traditional algebraic topology tools?

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Short answer: no. Such a new proof would be a notable event and you would have noticed the echo of it. –  Johannes Ebert Apr 25 '12 at 19:45

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