Suppose that we are talking about the linear map $$e^i\wedge:\text{Alt}(\otimes^pV^*)\to\text{Alt}(\otimes^{p+1}V^*),$$ which maps exterior $p$-forms to $(p+1)$-forms.

In my mind, this is the equivallent to the "exterior derivative" $d$ that arises in the context of differential geometry between differential forms living on the cotangent spaces of manifolds. That map is defined by $$d\omega(X_0,...,X_k)=\sum_i(-1)^iX_i(\omega(X_0,...,\hat X_i,...,X_k))+\sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_0,...,\hat X_i,...,\hat X_j,...,X_k),$$ where the $\hat X_i$ denotes the omiting of $X_i$ in the arguments etc.

My questions are the following:

  1. Is there a similar definition of the $e^i\wedge$ "exterior derivative", with respect to its vector arguments?
  2. Is the $e^i\wedge$ map unique? I mean, I suppose that there are lots of maps of this kind, since $i$ is unfixed.

1 Answer 1


You've just given a well-known construct in exterior algebra. (The analogy with the exterior derivative is really not right, because this construction will generalize to something linear over the $C^\infty$ functions, namely wedging with a fixed $1$-form on the manifold, whereas $d(f\omega) \ne f d\omega$ for general functions $f$.) The appearance of Lie brackets in the exterior derivative formula is, on the other hand, what's needed to make the resulting object a tensor field, namely (multi-)linear over $C^\infty$ functions in the sense that $d\omega(X_0,\dots,fX_i,\dots,X_k) = f d\omega(X_0,\dots,X_i,\dots,X_k)$. There is no such thing when we work in a fixed vector space, as scalars always pull out.

Given any vector $\alpha\in V^*$ and any vector $v\in V$, you can define two different linear maps: \begin{align*} e_\alpha\colon& \Lambda^k V^* \to \Lambda^{k+1} V^*, \quad\text{given by } e_\alpha(\tau) = \alpha\wedge\tau \\ \iota_v\colon& \Lambda^k V^* \to \Lambda^{k-1}V^*, \quad\text{given by } \iota_v(\tau)(v_1,\dots,v_{k-1}) = \tau(v,v_1,\dots,v_{k-1}). \end{align*} These are often called, respectively, exterior and interior product. Of course, you can write out the definition of the exterior product in terms of the value on $k+1$ vectors, with appropriate constants and signs on permutations. This is part of the standard definition in exterior algebra (if you think of it as a subspace of $\otimes^\bullet V^*$, rather than as a quotient space).

A further tangential comment: If you choose a basis $\{e_1,\dots,e_n\}$ for $V$ and the corresponding dual basis $\{\alpha_1,\dots,\alpha_n\}$ for $V^*$, then $$\iota_{v_i}\big(e_{\alpha_i} \tau\big) = \tau \qquad\text{and}\qquad e_{\alpha_i}\big(\iota_{v_i}\tau\big) = \tau.$$ (Depending on numeric conventions, perhaps some constant may appear.)

  • $\begingroup$ I know that we are discouraged from commenting things like "Thanks" etc, but thank you very much for this clear explanation! $\endgroup$
    – Ozz
    Sep 20, 2018 at 23:07
  • 1
    $\begingroup$ You're welcome, @GK. I hope you enjoy contributing to the site in various ways. $\endgroup$ Sep 20, 2018 at 23:12
  • $\begingroup$ One more question: I am thinking about the quantity $\iota_{v_i}e_{\alpha_i}+e_{\alpha_i}\iota_{v_i}$. This will give $2\tau$ when it acts on a form $\tau$ and it just reminds me of the Cartan formula for the Lie derivative. Is there any comment on this? I mean does the aforementioned quantity represent anything equivallent to the Lie derivative? $\endgroup$
    – Ozz
    Sep 21, 2018 at 21:08
  • $\begingroup$ Well, not so far as I can tell, since it's just a scalar multiple of the identity. The Cartan formula for the Lie derivative is really a chain homotopy formula, saying the Lie derivative is chain homotopic to the zero map. It's once again very hard to analogize things in a static vector space with things on a manifold with the structure of $d$ and vector fields. $\endgroup$ Sep 21, 2018 at 22:03

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