# Formular for interior product

1. Jul 13, 2014

In Nakahara's book, the interior product is defined like this :
$i_{x} \omega = \frac{1}{r!} \sum\limits_{s=1}^r X^{\mu_{s}} \omega_{\mu_{1}...\mu_{s}...\mu_{r}}(-1)^{s-1}dx^{\mu_{1}} \wedge ...\wedge dx^{u_{s}} \wedge...\wedge dx^{\mu_{r}}$

Can someone give me please a concret example of this? I can't make sense out of it. For example, how does this look explicit with $i_{e_{x}}(dx \wedge dy) = dy$?

Greets

2. Jul 14, 2014

### WWGD

I think the Wiki article is good:
http://en.wikipedia.org/wiki/Interior_product

$( \iota_X\omega )(X_1,\ldots,X_{p-1})=\omega(X,X_1,\ldots,X_{p-1})$
for any vector fields $X1,..., Xp−1.$

The interior product is the unique antiderivation of degree −1 on the exterior algebra such that on one-forms α
$\displaystyle\iota_X \alpha = \alpha(X) = \langle \alpha,X \rangle,$

Last edited: Jul 14, 2014
3. Jul 15, 2014

### Geometry_dude

One should view it as the generalization of matrix multiplication to higher rank tensors. WWGD has already given the appropriate formulas.
EDIT:
To compute
$$\partial_1 \cdot (d x^1 \wedge d x^2)$$
where $X \cdot \equiv i_{X}$ in your notation, I recommend to use
$$X \cdot (\alpha \wedge \beta) = (X \cdot \alpha )\wedge \beta + (-1)^k \, \alpha \wedge (X \cdot \beta)$$
for $\alpha$ a $k$-form and $\beta$ an $l$-form, which is a really useful formula. Try it!
As you asked to use the definition
$$\omega= d x^1 \wedge d x^2 = \frac{1}{2} \omega_{ij} \, d x^i \wedge d x^j = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
$$X = \partial_1 = X^i \, \partial_i =\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
hence
$$X \cdot \omega = \frac{1}{2} (X^i \omega_{ij} \, d x^j - X^i \omega_{ji} \, d x^j) = X^i \omega_{ij} \, d x^j = \begin{pmatrix} 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = d x^2$$
Here you also see what I meant with it just being ordinary matrix multiplication.

Last edited: Jul 15, 2014
4. Jul 21, 2014

Hey, sry I didn't notice your answer. Thank you very much. I think about it. I really have to get used to this notation.

Greets

5. Jul 22, 2014

### Geometry_dude

Don't be discouraged. It's all just vectors, covectors and matricies multiplied from the right and left. If you have a real vector space $V$ and a basis $e_1, \dots, e_n$, then
$$v= v^i \, e_i .$$
If you have a functional $\phi$, that is a map that takes a vector and spits out a real number, then you can write it as a covector
$$\phi = \phi_i \, e^i$$
where $e^i$ is the dual vector such that
$$e^i \cdot e_j = \delta^i_j \, .$$
This is a consequence of Riesz representation theorem in the special case of a finite dimensional inner product space, which is necessarily a hilbert space.
This is how it works:
$$\phi(v) \equiv \phi \cdot v = \phi_i \, e^i \cdot v^j \, e_j = \phi_i \, v^j \, (e^i \cdot e_j) = \phi_i \, v^j \, \delta^i_j = \phi_i \, v^i \, .$$
Now, a "normal" matrix $A$ (in your vector space $V$) is something that takes a vector and spits out a vector, so
$$A = A^i{}_j \, e_i \otimes e^j \, .$$
In a similiar manner, you can construct maps that take a vector and spit out a covector, maps that take a covector and spit out a vector, etc.
This differential geometry stuff is just $V$ being the tangent space and
$$e_i = \partial_i := \frac{\partial}{\partial x^i} \quad, \quad e^i = d x^i \, .$$
The "interior product" is just multiplication.

6. Jul 22, 2014

Hi, thx, yea that really helped, I have to think more in terms of matrices. So when I have a form with coefficients like $w_{\mu v}$ then this would be a matrix full of forms and this would take a vector(field) as input and a 1-form as output?. But I'm not sure what the role of the tensor product here is. I know that the forms are defined via the tensor product and I know how to take the tensor product of two vectors to get a "tensor matrix", but how does this mix all together?

7. Jul 22, 2014

### WWGD

Actually, the product in the subalgebra of differential forms ( a subalgebra of the (graded) exterior algebra $V^{n\otimes}$ of a vector space V, where the product on tensors is actually the tensor product) is the wedge product. The differential forms are the subalgebra of alternating tensors, and the product in the (also graded, by the degree) subalgebra is the wedge product. And the only matrix I know of with forms as entries is the connection form, tho maybe there is some other one out there I don't know of.

Last edited: Jul 22, 2014
8. Jul 22, 2014

### Geometry_dude

You have to go back to linear algebra. A matrix is not a collection of numbers written in rows and columns, a matrix is an "abstract" linear map that takes in a vector and spits out a vector or equivalently, that takes a vector from the right and a covector from the left and gives you a number. The collection of numbers written in rows and columns is just a representation of a matrix. Just like you can write vectors with respect to different bases you can write matricies with respect to different bases. The notation
$$A= A^i{}_j \, e_i \otimes e^j$$
just makes this explicit. On a conceptual level, you get the components by taking your "abstract" matrix $A$ and computing
$$A^i{}_j := A(e^i, e_j) \equiv e^i \cdot A \cdot e_j \, \in \mathbb R$$

Analogously, in differential geometry, you have an "abstract" $k$-form $\omega$ and a basis $(\partial_\mu)_{\mu=0}^n$, and you define
$$\omega_{\mu_1 \dots \mu_k} := \omega( \partial_{\mu_1}, \dots, \partial_{\mu_k}) \equiv \partial_{\mu_k} \cdot ( \partial_{\mu_{k-1}} \cdot ( \dots (\partial_{\mu_1} \cdot \omega)))\cdots$$
such that you can write $\omega$ (at least locally) as
$$\omega = \omega_{\mu_1 \dots \mu_k} \, d x^{\mu_1} \otimes \dots \otimes d x^{\mu_k} = \omega_{[\mu_1 \dots \mu_k]} \, d x^{\mu_1} \otimes \dots \otimes d x^{\mu_k}$$
$$= \omega_{\mu_1 \dots \mu_k} \, d x^{[\mu_1} \otimes \dots \otimes d x^{\mu_k]} = \frac{1}{k!}\omega_{\mu_1 \dots \mu_k} \, d x^{\mu_1} \wedge \dots \wedge d x^{\mu_k} \, .$$
So the $\omega_{\mu_1 \dots \mu_k}$ are just (local) functions on your manifold. This also shows how the different tensor products relate.

So if you take a vector field $X=X^\mu \, \partial_\mu$, then
$$X \cdot \omega = (X^\nu \, \partial_\nu ) \cdot (\omega_{\mu_1 \dots \mu_k} \, d x^{\mu_1} \otimes \dots \otimes d x^{\mu_k}) = X^\nu \omega_{\mu_1 \dots \mu_k} (\partial_\nu \cdot d x^{\mu_1}) \otimes d x^{\mu_2} \otimes \dots \otimes d x^{\mu_k}$$
$$= X^\nu \omega_{\mu_1 \dots \mu_k} \delta_\nu ^{\mu_1} \, d x^{\mu_2} \otimes \dots \otimes d x^{\mu_k} = \omega_{\mu_1 \dots \mu_k} \, X^{\mu_1} d x^{\mu_2} \otimes \dots \otimes d x^{\mu_k} = \frac{1}{(k-1)! } \omega_{\mu_1 \dots \mu_k} \, X^{\mu_1} d x^{\mu_2} \wedge \dots \wedge d x^{\mu_k}$$
So all you do is "plugging" $X$ into the first "slot". Nothing complicated here if viewed in the right light.

Last edited: Jul 22, 2014
9. Jul 22, 2014

### WWGD

Nice, Geometry Dude, you cleared up my confusion on the use of tensors with "differential forms" (I thought the algebra product on differential forms is the wedge product and not the tensor product). These are actually tensors and not really forms. After "Alternizing" ( transforming the tensors into being alternating) , the tensors become forms, and then we use the wedge on them.

10. Jul 22, 2014

### Geometry_dude

Yes, mathematicians tend to make everything complicated by talking about isomorphisms and operators all the time without actually writing down what they're doing. This can become really confusing when you're trying to use this stuff. This calculation shows that the wedge product is just an anti-symmetrized tensor product. You can define a symmetrized product $\vee$ analogously. Took me a while to figure this stuff out myself.

11. Jul 22, 2014

### WWGD

As a mathematician ( in training ), I think the problem is often just that of mathematicians doing a poor job of teaching mathematics, or not knowing the subject well-enough themselves (people are hired , not because they are good teachers, but because the schools want the prestige of the research done by hirees.)

Still, to be fair, it is difficult to simplify and be clear when using the maze of indices, subscripts and superscripts and just nasty notation that seems intrinsic to differential geometry.

12. Jul 22, 2014

Ok thanks really for the afford Geometry_dude. I think you mean it like this:

When you take the general definitions
$\alpha^{1} \wedge \beta^{1} = \alpha \otimes \beta - \beta \otimes \alpha$(for one forms)
$\alpha \wedge \beta(X,W) =\alpha(X)\beta(W)-\beta(X)\alpha(W)$

$X=y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}$

$\omega=3dx \wedge dy -(14zx+2) dx \wedge dz$

$\omega = 3(dx\otimes dy - dy \otimes dx) - 14zx (dx\otimes dz - dz\otimes dx)$

$\omega = 3(dx(X)\otimes dy - dy(X) \otimes dx) - 14zx (dx(X)\otimes dz - dz(X)\otimes dx)$

$\omega = 3[dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dy - dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) \otimes dx] - 14zx [dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dz - dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dx]$

And then use the orthogonality relation $\delta^{i}_{j}$. But thinking about this in terms of matrix multiplication is easier or not? How would this look like with matrix multiplication?

@WWGD: What I meant by matrix of forms was about the post of Geometry_dude with $\omega = dx^{1} \wedge dx^{2}= \frac{1}{2}\omega_{ij}dx^{i}\wedge dx^{j} = \begin{pmatrix} \omega_{11}dx^{1}\wedge dx^{1} & \omega_{12}dx^{1}\wedge dx^{2} \\ \omega_{21}dx^{2}\wedge dx^{1} & \omega_{22}dx^{2}\wedge dx^{2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ I think this is what you meant Geometry_dude right?