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Let $V$ be an $n$-dimensional vector space.

The exterior algebra $\Lambda V$ of $V$ is the direct sum of the exterior powers $\Lambda^kV$. It comes with a product (called the exterior product) which is bilinear, alternating and anticommutative. The dimension of $\Lambda V$ is $2^n$.

(For physics buffs, the exterior algebra is an example of a supersymmetric and supercommutative algebra.)

The exterior algebra is a geometric algebra with trivial quadratic form (I think). It is a quotient of the tensor algebra of $V$ (by the two-sided ideal generated by set $\{v \otimes v\ |\ v\in V\}$).

The exterior algebra $\Lambda\textbf{R}^1$ of $\textbf{R}^1$ is isomorphic to the dual numbers.

What is the exterior algebra $\Lambda\textbf{R}^2$ of $\textbf{R}^2$ (as in, through which isomorphic objects can I understand it, specially for the purpose of doing explicit computations)? Is it related to $\Lambda\textbf{C}$? Does it admit a matrix representation, like $\Lambda\textbf{R}^1$?

(If this question is too low-powered please do migrate it / close it.)

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    $\begingroup$ Exterior algebras always admit a matrix representation. After choosing a basis of $V$, you derive from it a basis for $\Lambda V$ (which is a $2^n$ dimensional vector space). Since $\Lambda V$ acts on $\Lambda V$ by left multiplication, there is a copy of $\Lambda V$ sitting in the algebra of linear maps $\mathcal{L}(\Lambda V, \Lambda V)$, which in your bases can be identified with the set of $2^n \times 2^n$ dimensional matrices. For example, for $V = \mathbf{R}^2$, you can let your basis of $\Lambda V$ be $\{1, dx, dy, dx\wedge dy\}$. Then the corresponding matrix operations are: $\endgroup$
    – Willie Wong
    Jul 6, 2017 at 17:54
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    $\begingroup$ \begin{align} 1 &\implies \begin{pmatrix} 1 \\ & 1 \\ & & 1 \\ & & & 1\end{pmatrix} \\ dx &\implies \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \\ dy & \implies \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 & -1 & 0 & 0\end{pmatrix} \\ dx \wedge dy & \implies \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 & 0 & 0 & 0\end{pmatrix} \end{align} with all omitted entries being 0. $\endgroup$
    – Willie Wong
    Jul 6, 2017 at 17:59
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    $\begingroup$ It is a finite dimensional local selfinjective algebra and thus a quiver algebra. It is an easy excerse to write down quiver and relations (which is imo much better than as a matrix algebra). This is for example an excerse in the book by auslander reiten smalo. Having the algebra as a quiver algebra, one can for example easily answer when it is symmetric algebra. I dont know when it is a Hopf algebra. $\endgroup$
    – Mare
    Jul 6, 2017 at 18:10
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    $\begingroup$ @Mare: it is always a Hopf algebra. $\endgroup$
    – M.G.
    Jul 6, 2017 at 18:48
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    $\begingroup$ @July: It is a Hopf algebra in the category of super-vector spaces. But it is not generally a Hopf algebra "on the nose" (i.e., without the Koszul sign). For example, the exterior algebra of a $1$-dimensional vector space over a field $k$ is isomorphic to $k\left[x\right]/\left(x^2\right)$, which has no bialgebra structure unless $k$ has characteristic $2$. $\endgroup$
    – darij grinberg
    Jul 6, 2017 at 21:30

1 Answer 1

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The exterior algebra $\Lambda \mathbb{R}^2$ is a real vector space of dimension 4 with basis $1, e_1, e_2, e_1 \wedge e_2$. So its every element is a unique linear combination of these basis elements, say $a_1 \cdot 1 + a_2 e_1 + a_3 e_2 + a_4 e_1 \wedge e_2$, for real numbers $a_1, a_2, a_3, a_4$, which can be chosen arbitrarily. The multiplication operation is written $\wedge$, is associative and has $1$ as identity, and has $e_1 \wedge e_1 = 0$ and $e_2 \wedge e_2=0$, $e_1 \wedge e_2$ is the basis element by that name, and $e_2 \wedge e_1=-e_1 \wedge e_2$. That gives you the complete multiplication table, by associativity. Please comment if this answer is not sufficient.

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    $\begingroup$ You mean the exterior algebra $\Lambda\mathbb{R}^2$ (if you only take the second exterior power, you won't get an algebra). Also, you need to specify that $e_2\wedge e_1 = -(e_1\wedge e_2)$ to complete the multiplication table. $\endgroup$
    – Tobias Kildetoft
    Jul 7, 2017 at 13:16
  • $\begingroup$ @TobiasKildetoft: thanks, fixed. $\endgroup$
    – Ben McKay
    Jul 7, 2017 at 13:57
  • $\begingroup$ Is $\Lambda \textbf{R}^2$ isomorphic to any object one may encounter in elementary math? (Like $\Lambda \textbf{R}^2$ is isomorphic to the dual numbers, which form a unital commutative associative algebra.) What are some other properties of $\Lambda \textbf{R}^2$, as an algebra? $\endgroup$
    – étale-cohomology
    Jul 24, 2017 at 11:58
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    $\begingroup$ @étale-cohomology: No, it is not isomorphic to the dual numbers. They have dimension 2. It has dimension 4. The even part of $\Lambda \mathbb{R}^2$ is the dual numbers: $\Lambda^0 \mathbb{R}^2 \oplus \Lambda^2 \mathbb{R}^2$. The algebra $\Lambda \mathbb{R}^2$ is not commutative, but is extremely close to being commutative (called "super-commutative"), since it splits (as vector space) into the direct sum of its even part, a commutative algebra, and then its odd part, $\Lambda^1 \mathbb{R}^2$, with commutativity of anything even and anything, but anti-commutativity of any two odd things. $\endgroup$
    – Ben McKay
    Jul 24, 2017 at 16:37
  • $\begingroup$ Thanks for the answer! And I meant that $\Lambda \textbf{R}^1$ is isomorphic to the dual numbers. $\endgroup$
    – étale-cohomology
    Jul 25, 2017 at 17:03

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