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Let $V$ denote an $n$-dimensional real vector space, and $\wedge^n$ denote the $n$-fold exterior product. What is the isomorphism between $\wedge^n(V)$ and $\mathbb{R}$?

In the book Introduction to Manifolds, by Loring W. Tu, 3rd edition, Chapter: Orientation of Manifolds, it says:

A linear isomorphism $\wedge^n(V)\cong\mathbb{R}$, identifies the set of non-zero $n$-covectors on $V$ with $\mathbb{R} \setminus \{0\}$".

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I think it's helpful if you specify what your symbols mean. I guess that $V$ denotes an $n$-dimensional real vector space and $\wedge^n$ the $n$-fold exterior product. – Hans Aug 21 '12 at 19:24
Let $\eta \in \bigwedge^n V \setminus \{0\}$, then $\lambda \mapsto \lambda \eta$ is a linear isomorphism $\mathbb R \to \bigwedge^n V$. – martini Aug 21 '12 at 19:26
@Filippo yes exactly... – Un Chien Andalou Aug 21 '12 at 19:27
It's not exactly correct to say "the isomorphism." There are many. One must choose a basis for the top exterior power, i.e. a non-zero vector, for example by choosing a basis for $V$. – Keenan Kidwell Aug 21 '12 at 19:33
up vote 6 down vote accepted

Note that you assume that $V$ is of dimension $n$. The $n$-fold outer product of $V$ can now be described as follows: Choose a basis $v_1,\dots,v_n$ of $V$. The outer product is generated (as a vector space) by expressions of the form $v^1\wedge\dots\wedge v^n$ where $v^1,\dots,v^n\in\{v_1,\dots,v_n\}$, where one such expression $v^1\wedge\dots\wedge v^n$ is the negative of $w^1\wedge\dots\wedge w^n$ if $w^1\wedge\dots\wedge w^n$ is obtained from $v^1\wedge\dots\wedge v^n$ by exchanging one component $v^i$ and another component $v^j$. This implies that $v^1\wedge\dots\wedge v^n=0$ if there are distinct $i$ and $j$ with $v^i=v^j$.

It follows that $v_1\wedge\dots\wedge v_n$ and $-v_1\wedge\dots\wedge v_n$ are the only non-zero generators, and they are linearly dependent (obviously). Hence the outer product is generated by $v_1\wedge\dots\wedge v_n$. Mapping this generator to $1$ extends to your desired isomorphism. As pointed out in the comments, there are many isomorphisms, and the one pointed out here depends on the choice of the basis $v_1,\dots,v_n$.

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Minor nitpick: The generators are obviously linearly dependent. – Johannes Kloos Aug 21 '12 at 19:49
Indeed. I edited this. – Stefan Geschke Aug 21 '12 at 21:02

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