Determinants, Permutations and the Lie Algebra of SL(n)

Here is an old classic from linear algebra: given an $ n\times n$ matrix $ A = (a_{ij})$, the determinant of $ A$ can be calculated using the permuation formula for the determinant:

$ \det(A) = \sum_{\sigma\in S_n} (-1)^\sigma a_{1\sigma(1)}\cdots a_{n\sigma(n)}$.

Here $ S_n$ denotes the permutation group on $ n$ symbols and $ (-1)^\sigma$ denotes the sign of the permutation $ \sigma$. The computation of the determinant is much more easily done via the 'minor expansion method', but this just reduces to the above formula. Now, typically we would never use the above formula to calculate the determinant, but that doesn't mean it isn't useful!

The Lie Algebra of an Algebraic Group

In fact, computating the Lie algebra of $ \mathrm{SL}_n$ over a fixed commutative ring $ k$ happens to be a quick and interesting application of the above determinant formula. First, let us recall the definition of the Lie algebra, or at least one of the many equivalent definitions. Consider the algebra of dual numbers, $ k[\tau]/(\tau^2)$. Each element in this algebra can be written uniquely as $ a + b\tau$: this gives a $ k$-algebra map $ C: k[\tau]/(\tau^2)\to k$ by sending $ a + b\tau$ to $ a$.

The Lie algebra of an algebraic group $ G$ defined over a ring $ k$ is then the set of $ k$-algebra homomorphisms $ T:k[G]\to k[\tau]/(\tau^2)$ that satisfy $ C\circ T = \epsilon$ where $ \epsilon:k[G]\to k$ is the counit: the morphism that gives the identity element of $ G$. I'll quickly remark that one may consider all $ T:k[G]\to k[\tau]/(\tau^2)$ that gives the bundle of Lie algebras over $ G(k)$.

One may easily compute that for $ \mathrm{GL}_n$, its Lie algebra is the set of all $ I + M\tau$ where $ I$ is the $ n\times n$ identity and $ M$ is an arbitrary matrix. The $ I$ is a restatement that $ C\circ T = \varepsilon$ and the $ M$ is an arbitrary matrix—this is because every such element $ I + M\tau$ is actually invertible (what is its inverse???). Since the $ I$ is always the same we may as well just consider the Lie algebra of any group to be the set of all $ M$ such that $ I + M\tau$ corresponds to an element in the originally-defined Lie algebra. It is not hard to show then that the Lie algebra of any group is actually closed under the Lie bracket, making our definition well-defined.

The Lie Algebra of $ \mathrm{SL}_n$

In order to compute the Lie algbera of $ \mathrm{SL}_n$, we just have to recall that $ \mathrm{SL}_n$ is the group of all matrices whose determinant is one. Thus, for the Lie algebra, an element $ I + M\tau$ will be in the Lie algebra of $ \mathrm{SL}_n$ if and only if its determinant is one. We consider $ I + M\tau$ as a single matrix whose entries are of the form $ \delta_{ij} + m_{ij}\tau$ where $ m_{ij}$ are the entries of $ M$ and $ \delta_{ij} = 1$ if and only if $ i=j$ and vanishes otherwise. Of course then the determinant will have values in the ring $ k[\tau]/(\tau^2)$.

Now we come back to the handy dandy formula

$ \det(A) = \sum_{\sigma\in S_n} (-1)^\sigma a_{1\sigma(1)}\cdots a_{n\sigma(n)}$.

If we plug in the matrix $ I + M\tau$, we will get a long polynomial in $ \tau$ that looks like $ a_0 + a_1\tau + a_2\tau^2 + \cdots + a_n\tau^n$. Of course, since we are actually computing in the ring $ k[\tau]/(\tau^2)$, the order two and higher terms just are zero. Now we consider the matrix $ I + M\tau$. For most permutations, the corresponding term in the above sum will actually just be zero. Actually, if in the term for $ \sigma$ there are at least two terms of the form $ 0 + a_{ij}\tau$, then the lowest power of $ \tau$ will be at least squared, so we can just ignore all terms of this kind. However, if there is at least one term not of the form $ (1 + a_{ii}\tau)$ then there are at least two, because $ \sigma$ is a permutation after all!

Hence, the only nontrivial term in the above computation of the determinant will be when $ \sigma$ is the identity permutation, so that $ \det(I + M\tau) = 1 + \mathrm{trace}(M)\tau$. But we require this element to be one, so concluding, $ I + M\tau$ is in the Lie algebra of $ \mathrm{SL}_n$ if and only if $ \mathrm{trace}(M) = 0$!

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