Definition of "interior derivative" and "exterior derivative"? 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!
 A: *

*
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
$ι_vω(X_2,…,X_n)=ω(v,X_2,…,X_n)$

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