Suppose we have a $N$-dimensional vector space $V$, equipped with a scalar product $\langle\cdot{ , }\cdot\rangle$ and let $\{e_1,\dots,e_N\}$ be a orthonormal basis.

I've seen that there are different ways to define $\bigwedge^k V$, the exterior $k$-power of $V$, but the most operative one (for my interests) is:

$$\bigwedge ^k V=\text{Span}\{e_{i_1}\wedge \dots \wedge e_{i_{k}}|i_1<\dots<i_k,\, 1\leq k\leq N \},$$ where we can think to $\wedge$ as a anticommutative symbol of multiplication. We can define a scalar product on each $\bigwedge^k V$, requiring that

$$\langle { e_{i_1}\wedge \dots \wedge e_{i_{k}}, } e_{i_1}\wedge \dots \wedge e_{i_{k}} \rangle=1, \\ \langle { e_{i_1}\wedge \dots \wedge e_{i_{k}}, } e_{j_1}\wedge \dots \wedge e_{j_{k}} \rangle=0 \text { if } e_{i_1}\wedge \dots \wedge e_{i_{k}}\neq \pm e_{j_1}\wedge \dots \wedge e_{j_{k}}. $$

Now, $\bigwedge^{N-1}V$ has the same dimension as $V$, its basis is given by $$\{e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N |1\leq i \leq N \}.$$ To each vector $e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N$ it is associated the vector $e_i\in V.$ My professor has said that this kind of correspondence is "equivalent" to say that $e_i$ is orthogonal to $e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N$, but I don't understand in which sense they are orthogonal. Can we identify ${e_1\wedge\dots \wedge \hat{e}_i\wedge \dots \wedge e_N}^\perp$ with $e_i$? If so, how?

  • $\begingroup$ Are you using \Lambda for \bigwedge? $\endgroup$
    – Asaf Karagila
    Commented Oct 19, 2015 at 10:21
  • $\begingroup$ Yes, isn't it the standard notation? $\endgroup$
    – batman
    Commented Oct 19, 2015 at 11:24
  • $\begingroup$ I don't know about the notation, I was talking about the LaTeX. $\endgroup$
    – Asaf Karagila
    Commented Oct 19, 2015 at 11:28
  • $\begingroup$ Ok, writing $\Lambda^{N-1}V$ I mean $V\wedge V\wedge V\wedge \dots\wedge V$ $N-1$ times, as a subspace of $V\otimes V\otimes \dots \otimes V$ $N-1$ times. $\endgroup$
    – batman
    Commented Oct 19, 2015 at 11:35
  • 1
    $\begingroup$ I understand that. I meant that you should write $\bigwedge^{N-1}$ and not $\Lambda^{N-1}$. $\endgroup$
    – Asaf Karagila
    Commented Oct 19, 2015 at 11:37

2 Answers 2


The easiest way to answer your question is first bring it in an 'equivalent' form and then answer it. Of course you might still argue whether the the reversed form is really equivalent...

The modified question is:

Can we identify $e_1 \wedge \cdots \wedge \hat{e}_i \wedge \cdots \wedge e_N$ with $e_i^\perp$?

(So I only moved the $\perp$ to the other side of the $=$-sign)

The answer is: yes we can! Note that $e_i^\perp$ is a hyperplane in $V$ and the indentification is part of a much more general identification between non-zero elements of $\bigwedge^{N-1}V$ and hyperplanes in $V$ (that is: $(N-1)$-dimensional subspaces of $V$): just send any element $v_1 \wedge \cdots \wedge v_{N-1}$ to the hyperplane spanned by the vectors $v_1, \ldots, v_{N-1}$! (Note that if they don't span a hyperplane, the element $v_1 \wedge \cdots \wedge v_{N-1}$ equaled zero in the first place!)

Now what is not so obvious from the way I wrote the identification from elements of $\bigwedge^{N-1}V$ to planes is that non-zero elements of $\bigwedge^{N-1}V$ that 'look' different but define the same hyperplane in $V$ are in fact already the same element in $\bigwedge^{N-1}V$ up to a scalar factor. However all the nice extra structure on $\bigwedge^{N-1}V$ you defined in your post can be very helpful in showing that.


Hint: Look at the Hodge Star isomorphism

$$\begin{align}\star: \Lambda^{k}V &\to \Lambda ^{n-k}V\\e_{i_1} \wedge \ldots \wedge e_{i_k} &\mapsto (-1)^{\sigma} e_{j_{1}}\wedge \ldots \wedge e_{j_{n-k}}\end{align} $$

where $i_1 < \ldots < i_k$, $j_1 < \ldots < j_{n-k}$, $(i_1,\ldots, i_k,j_1,\ldots,j_{n-k} )$ is a permutation of $(1,2,\ldots, n)$ and $\sigma$ is $0$ or $1$ whether the permutation is even or odd, respectively.

Now taking $k =1$ and $\dim V = n$ we notice that $n=\dim V = \dim \Lambda^1 V = \dim \Lambda^{n-1} = \binom{n}{1}$. Then it makes sense to talk about a correspondence here.

$$\begin{align}\star: \Lambda^{1}V &\to \Lambda ^{n-1}V\\e_{i_1} &\mapsto (-1)^{\sigma} e_{j_{1}}\wedge \ldots \wedge e_{j_{n-1}}\end{align} $$

Finally there is a correspondence $V \simeq \Lambda^{1}V$ through $e_{i_1}(\varepsilon_j) = \delta_{i_1}^{j}$, (Kronecker delta).

  • $\begingroup$ How can I show that the Hodge Star operator is independent of choice of basis ? $\endgroup$
    – user371231
    Commented Apr 11, 2018 at 9:59
  • $\begingroup$ The Hodge star is independent of a basis but it does depend on the choice of inner product (and volume form). Basis free definition: a\wedge *b = (a,b) w for a,b k-vectors and w the volume form $\endgroup$
    – Callum
    Commented Nov 10, 2020 at 17:53

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