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!


1 Answer 1


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 $\{1, \ldots, n\}$.

  • $\begingroup$ 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. $\endgroup$
    – Joppy
    Jan 18, 2021 at 9:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .