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In Willie Wong's reply to one question, he used some concepts: "interior derivative" of a differential form and "exterior derivative" of a scalar function on $\mathbb{R}^3$.

For "exterior derivative" of a scalar function on $\mathbb{R}^3$, I think it means the exterior derivative of the scalar function viewed as a 0-form.

For "interior derivative", I am not able to find the definition from elsewhere. Here is his original text:

Let $\omega$ be a volume form on some manifold $M$. (So if $M$ has $n$-dimensions, $\omega$ is a differentiable $n$-form.) Via the volume form we can define the notion of volume, and the notion of an integral in the usual way. (I assume you are familiar with that already.) Then the interior derivative $\iota_v\omega$, which is the $n-1$-form defined by

$$ \iota_v\omega(X_2,\ldots,X_n) = \omega(v,X_2,\ldots,X_n) $$

for $v$ a vector field on $M$, is a differentiable form of the top degree when restricted to any $n-1$-dimensional submanifold.

Must an interior derivative of a differential form be specified relative to a vector field?

May I have some clue and references here? Thanks in advance!

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I think this is what is sometimes called interior multiplication. – Dylan Moreland Dec 5 '11 at 0:43
@DylanMoreland: Thanks! Are exterior derivative, exterior product and exterior multiplication the same thing? – Tim Dec 5 '11 at 0:49
"Exterior multiplication" is not something I've heard before. The derivative and product are not the same, as far as I can tell. The product is just the ring operation on $\bigwedge V$ and the derivative is an operator (a graded antiderivation, or something) on $\bigwedge T^*M$. – Dylan Moreland Dec 5 '11 at 1:09
@DylanMoreland: Thanks! I wonder when Wiki says the interior product is "named in opposition to the exterior product", if it means that interior product and exterior product are contrast to each other more than just in their names? Is interior product more opposite to exterior derivative? – Tim Dec 5 '11 at 2:43
I've never really thought of them as opposite, but then again my differential geometry is pretty bad. I guess this could be true in the sense that one operation raises the rank of the tensor and the other lowers it. I must admit that I haven't really read the question carefully -- I'll try to do so later. – Dylan Moreland Dec 5 '11 at 2:45
  1. Yes, the interior derivative is always specified relative to a vector field. It is very simple. In fact, you quote a complete definition in your question.

Then the interior derivative $ι_vω$, which is the n−1-form defined by


That's really all there is to it. You take some n-form (with n>0) and to figure out how it acts on some other set of vector fields, you simply "insert" the vector field you took the interior product with into the arguments. Look at the above expression, it's simpler than words can describe. It shows how a n-1 form $ι_vω$ acts on a series of vector fields $X_2,…,X_n$.

Note, however, that this doesn't work on 0-forms. You can see this one of two ways. First, it takes n forms to n-1 forms, and there's no such thing as -1 forms. Second, if you look at how its defined above, there's nowhere to "insert" the vector field as an argument, because a 0-form takes 0 arguments. So the interior derivative of a 0-form (a function) is just always 0.

2. As Dylan Pointed out, the interior product is the same as the interior derivative. See the Wikipedia article.

3. The exterior product is not the same as the exterior derivative. The exterior product is just another name for the wedge product. The exterior derivative is the $d$ operator.

There is an close relationship between exterior derivatives, interior derivatives, and lie derivatives, see Cartan's identity:

$\mathcal L_X\omega = \mathrm d (\iota_X \omega) + \iota_X \mathrm d\omega$

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