Let us suppose we have a (finite-dimensional) vector space $X$ and let $\bigwedge^k X$ be its exterior product of order $k$. Then $a_1 \wedge a_2 \wedge \dots \wedge a_k \in \bigwedge^k X$ denotes the exterior product of $a_1, a_2, \dots, a_k \in X$.

Question: What is the relation between $\bigwedge^4 X$ and $\bigwedge^2 ( \bigwedge^2 X )$?

For example, suppose we have two exterior products $a \wedge b$ and $c \wedge d$. What is the relation between $(a \wedge b) \wedge (c \wedge d)$ and $a \wedge b \wedge c \wedge d$?

It seems that in $(a \wedge b) \wedge (c \wedge d)$ obeys the (anti)symmetries

$(a \wedge b) \wedge (c \wedge d) = -(b \wedge a) \wedge (c \wedge d)$

$(a \wedge b) \wedge (c \wedge d) = -(a \wedge b) \wedge (d \wedge c)$

$(a \wedge b) \wedge (c \wedge d) = (c \wedge d) \wedge (a \wedge b)$

but I do not see any antisymmetry such as

$(a \wedge b) \wedge (c \wedge d) = -(a \wedge c) \wedge (b \wedge d)$

I suspect I am getting confused about the notation. Are there two different types of wedge products in the play?

  • 4
    $\begingroup$ The exterior product is associative, so they're the same. A hint that it is associative is that you can make sense of an expression like $a \wedge b \wedge c$. If it wasn't associative, this would be ambiguous notation and would not be well defined without adding parentheses accordingly. $\endgroup$ Commented Dec 25, 2020 at 16:51

2 Answers 2


The wedge product is usually defined to be associative. The fact that the expression $a\wedge c\wedge b\wedge d$ is well-defined already requires associativity, since otherwise we would need to specify the order of the wedge products.

That said, it is not the case that $\bigwedge^2(\bigwedge^2 X)\cong\bigwedge^4 X$. One easy way to see this is by dimension counting: Since $\dim(\bigwedge^kX)=\binom{\dim X}{k}$, we have $\dim(\bigwedge^2(\bigwedge^2 X))=3\binom{\dim X+1}{4}$ and $\dim(\bigwedge^4 X)=\binom{\dim X}{4}$, and these two dimensions are never equal unless both are zero. The issue is exactly as you suspect, there are two different wedge products, one introduced by each application of the $\bigwedge^2$ functor, and the "outer" one only acts on elements of $\bigwedge^2 X$.

  • $\begingroup$ Thanks for explaining. So there two different wedge products. I guess the best way to make sense of that is to keep in mind which spaces they are referring to. In the case $\bigwedge^2 (\bigwedge^2 X)$, or more generally $\bigwedge^m (\bigwedge^n X)$ the inner product is about X, and the outer product is about its exterior power. $\endgroup$
    – shuhalo
    Commented Dec 25, 2020 at 18:48

The exterior product defines a bilinear map $\wedge^i(V) \otimes \wedge^j(V) \to \wedge^{i+j}(V)$ for any $i, j$, and so in particular defines a bilinear map

$$\wedge^i(V) \otimes \wedge^i(V) \to \wedge^{2i}(V).$$

This map is alternating when $i$ is odd and symmetric when $i$ is even, so induces maps

$$S^2(\wedge^{2i}(V)) \to \wedge^{4i}(V)$$ $$\wedge^2(\wedge^{2i+1}(V)) \to \wedge^{4i+2}(V).$$

In particular, we do not get a natural map $\wedge^2(\wedge^2(V)) \to \wedge^4(V)$, but instead a natural map $S^2(\wedge^2(V)) \to \wedge^4(V)$. ("Natural" here means "natural in $V$" and so in particular these are all morphisms of $GL(V)$-representations.)

The phrasing of your question in the title is ambiguous because $(a \wedge b) \wedge (c \wedge d)$ would usually be interpreted in terms of the usual exterior product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^4(V)$ (which is symmetric), whereas you are asking about a different exterior product $\wedge^2(V) \otimes \wedge^2(V) \to \wedge^2(\wedge^2(V))$ (which is alternating). To be less ambiguous the two different exterior products involved should be notated differently.


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