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Does there exist a bi-Invariant metric on $SL_n(\mathbb{R})$. I tried to google a bit but I didn't find anything helpful.

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Doesn' t the negative of the Killing form do? –  Mariano Suárez-Alvarez Dec 20 '10 at 16:04
    
Ah: That gives a biinvariant semiRiemannian metric, rather. –  Mariano Suárez-Alvarez Dec 20 '10 at 16:06
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up vote 7 down vote accepted

A connected Lie group has a biinvariant metric iff it is isomorphic to the dirtect product of a compact one and a vector space---see Milnor's Curvatures of left invariant metrics. Such a factorization would carry over to the Lie algebras, etc.

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It's enough to show the answer is "no" for $SL_2(\mathbb{R})$, because $SL_2(\mathbb{R})$ is naturally embedded in $SL_n(\mathbb{R})$, and if $SL_n(\mathbb{R})$ has a biinvariant metric, then the induced metric on $SL_2(\mathbb{R})$ would be biinvariant.

So let's focus on $SL_2(\mathbb{R})$. Finding a biinvariant metric on $SL_2(\mathbb{R})$ is equivalent to finding an $Ad(SL_2(\mathbb{R})$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, so I'll show there is no such $Ad$ invariant metric.

Now, $\mathfrak{sl_2(\mathbb{R})}$ is 3 dimensional with basis $E_{12}, E_{21}, E_{11} - E_{22}$ where $E_{ij}$ denotes the matrix with a 1 in the $ij$ slot and a 0 elsewhere.

Consider $A = diag(2,1/2)$ ,the diagonal matrix with diagonal entries 2 and 1/2. This is clearly in $SL_2(\mathbb{R})$. Note then that $AE_{12}A^{-1} = 4 E_{12}$.

But if an inner product were $Ad$ invariant, it would have to preserve the length of $E_{12}$, giving a contradiction.

Thus, there is no $Ad$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, and hence there is no biinvariant metric on $SL_n(\mathbb{R})$.

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