# Exterior power of lie algebra representation in coordinates

Suppose I have a representation $$V = \mathbb{C}^3$$ of a group $$G$$ where elements are of the form $$M = \begin{bmatrix}m_1^1 & m_1^2 & m_1^3 \\ m_2^1 & m_2^2 & m_2^3 \\m_3^1 & m_3^2 & m_3^3 \end{bmatrix}$$ in some basis $$v_1,v_2,v_3$$. If I wish to compute the representation $$\Lambda^2 V \subset V \otimes V$$, I could look at the action $$g \cdot (v \wedge u) := (gv) \wedge (gu)$$ and see that I will get a matrix $$N$$ of minors of $$M$$, i.e. $$n_{i,j} =$$minor obtained by removing $$i$$th row and $$j$$th column.

My question is as follows: In the case of a representation of a Lie algebra $$\mathfrak{g}$$, is there an analogous interpretation when considering an exterior power of the representation?

As an example, using the Lie algebra action $$g(v \wedge u) := (gv) \wedge u + v \wedge (gv)$$, with basis $$v_1 \wedge v_2$$, $$v_1 \wedge v_3$$, and $$v_2 \wedge v_3$$, I obtain the following matrix for $$\begin{bmatrix}m_1^1+m_2^2 & m_2^3 & -m_1^3 \\ m_3^2 & m_1^1+m_3^3 & m_1^2 \\-m_3^1 & m_2^1 & m_2^2 + m_3^3 \end{bmatrix}$$. Any insight would be appreciated!

Consider a linear operator $$T \colon V \to V$$. This induces a linear operator $$\wedge^k T\colon \wedge^k V \to \wedge^k V$$ on the wedge power, and hence we get a map between Lie groups $$\varphi \colon \operatorname{GL}(V) \to \operatorname{GL}(\wedge^k V).$$ This map is polynomial (each coordinate on the right is a determinant of a matrix minor on the left), and therefore can be differentiated at the identity $$I \in \operatorname{GL}(V)$$ to a linear map $$d \varphi_I: \operatorname{End}(V) \to \operatorname{End}(\wedge^k V).$$ It is relatively straightforward (in principle) to compute $$d \varphi_I$$: let $$M = [m_i^j]$$ be a matrix of indeterminates, then the derivative should be the first-order term in the expansion $$d \varphi_I(I + tM) = I + t d\varphi_I(M) + t^2(\text{higher order terms...}).$$

For example, to compute the top-left entry of $$d \varphi_I(M)$$ in your example, we compute the determinant $$\det_{1, 1} (I + tM) = \det \begin{pmatrix} 1 + t m_1^1 & tm_1^2 \\ tm_2^1 & 1 + tm_2^2 \end{pmatrix} = 1 + t(m_1^1 + m_2^2) + t^2(...),$$ and therefore the result is $$m_1^1 + m_2^2$$. We could try another entry: $$\det_{1, 3} (I + tM) = \det \begin{pmatrix} t m_2^1 & 1 + tm_2^2 \\ tm_3^1 & tm_3^2 \end{pmatrix} = t(-m_3^1) + t^2(m_2^1 m_3^2 - m_3^1 m_2^2),$$ so the result is $$-m_3^1$$ (note that we don't expect a $$1+$$ at the start here, since $$\det_{1, 3}(I) = 0$$).

We can work this more carefully into a general rule. Say we are working with an $$n \times n$$ matrix $$M$$ and calculating in the $$k$$th exterior power. The matrix $$\varphi(M)$$ is indexed by pairs $$J, K$$ of $$k$$-element subsets of $$\{1, \ldots, n\}$$, where $$\varphi(M)_J^K = \det M_J^K$$, where $$M_J^K$$ is the $$k \times k$$ submatrix of $$M$$ with rows $$J$$ and columns $$K$$. This determinant can be written more explicitly as $$\det M_J^K = \sum_{\sigma : J \to K} (-1)^\sigma \prod_{j \in J} M_j^{\sigma(j)},$$ where $$\sigma$$ ranges over all bijections $$J \to K$$, and $$(-1)^\sigma$$ is $$(+1)$$ if the diagram of $$\sigma$$ has an even number of crossings, and $$(-1)$$ for an odd number of crossings. Now consider the determinant $$\det(I + tM)_J^K$$: the only summands that produce a $$t^1$$ term are either if:

1. $$\sigma(j) = j$$ for all $$j \in J$$, in which case $$J = K$$, and $$\det(I + tM)_J^J = 1 + t \operatorname{tr}(M_J^J) + t^2(...)$$. This is like the first example above.
2. $$\sigma(j) = j$$ for all but one $$j \in J$$, in which case we have $$|J \cap K| = k - 1$$, and $$\det(I + tM)_J^K = t (-1)^\sigma M_j^k + t^2(...)$$, where $$j$$ is the unique element of $$J \setminus K$$, $$k$$ is the unique element of $$K \setminus J$$, and $$\sigma$$ is the identity on $$J \cap K$$ and sends $$j$$ to $$k$$.

All other terms are zero: in particular, if $$|J \cap K| < k - 1$$ then $$d \varphi_I(M)_J^K = 0$$. To recap, we have

1. $$\operatorname{tr}(M_J^K)$$ if $$J = K$$,
2. $$(-1)^\sigma M_j^k$$ if $$|J \cap K| = k - 1$$, where $$\sigma: J \to K$$ fixes $$J \cap K$$ and $$j \in J \setminus K$$, and $$k \in K \setminus J$$,
3. $$0$$ if $$|J \cap K| < k - 1$$.

Finally, an example for $$\wedge^2 M$$ where $$M$$ is a $$4 \times 4$$ matrix. There are six two-element subsets of $$\{1, 2, 3, 4\}$$, and we get:

$$\begin{array}{@{} c c c c c c c @{}} & \{1, 2\} & \{1, 3\} & \{1, 4\} & \{2, 3\} & \{2, 4\} & \{3, 4\} \\ \{1, 2\}& m_1^1 + m_2^2 & m_2^3& m_2^4& -m_1^3& -m_1^4& 0\\ \{1, 3\}& m_3^2 & m_1^1 + m_3^3 & m_3^4& m_1^2& 0& -m_1^4\\ \{1, 4\}& m_4^2& m_4^3& m_1^1 + m_4^4 & 0 & m_1^2& m_1^3\\ \{2, 3\}& -m_3^1& m_2^1& 0& m_2^2 + m_3^3 & m_3^4& -m_2^4\\ \{2, 4\}& -m_4^1& 0 & m_2^1& m_4^3& m_2^2 + m_4^4 & m_2^3\\ \{3, 4\}& 0& -m_4^1& m_3^1& -m_4^2 & m_3^2& m_3^3 + m_4^4 \end{array}$$

The structure is rather simple, but the signs (as usual) make coordinate descriptions annoying. Even so, it is in some ways simpler than the matrix of minor determinants: most of the reason that Lie algebras are so useful is that representations become linear (rather than polynomial on a group).

The transpose-symmetry (ish) you can see in the matrix for $$\wedge^2 \mathbb{C}^4$$ is kind of special because this space is self-dual, in general there should be some kind of symmetry between the operators for $$\wedge^k \mathbb{C}^n$$ and $$\wedge^{n - k} \mathbb{C}^n$$, by complementing the sets indexing the rows/columns inside $$\{1, \ldots, n\}$$.

• I should also mention that with exterior algebras, signs only really appear once you try to rearrange everything into coordinates: you can go a long way (in terms of abstract theorems) by simply using $g \cdot (x\wedge y\wedge z) = gx \wedge y \wedge z + x \wedge gy \wedge z + x \wedge y \wedge gz$ and the like. Jan 18, 2021 at 9:00