# MTW: Exer. 8.12, pg. 215

1. Sep 2, 2010

### Living_Dog

In this problem the authors state:

I know that the first expression evaluates to the Kronecker delta in $$\alpha$$ and $$\beta$$ but the other expression is a 2x2 tensor. So how is the first (the identity tensor) the contraction (implying one less index) of the second, which is a 2nd rank tensor??

-LD

2. Sep 2, 2010

### phyzguy

It is confusing at first, but you need to understand the notation. The bold-faced quantities, omega and e, are one-forms and vectors, respectively. But their subscripts and superscripts do not refer to the components of omega and e, but are indices to denote which vector you are talking about. This is explained (briefly!) on page 51. Thus
$$\omega^{\alpha}$$
does not refer to the four components of a single one-form, but to a set of four one-forms. Then
$$<\omega^{\alpha},e_{\beta}>$$
is a set of 16 scalars, each of which is the innerproduct of one vector (one of the e's) and one one-form (one of the omegas). But the quantity:
$$\omega^{\alpha}\otimes e_{\beta}$$
is a set of 16 rank two tensors, each of which is the outer product of one e and one omega. So when I contract the rank-2 tensors on the RHS, I get a set of 16 scalar equations. Does this make sense?

3. Sep 2, 2010

### Living_Dog

The 2nd throws me off a bit. I understand both descriptions of these two expressions. What I am missing is how
$$<\bold{\omega}^{\alpha},\bold{e}_{\beta}>$$
is a contraction of
$$\bold{\omega}^{\alpha}\otimes \bold{e}_{\beta}$$.

E.g.

$$<\bold{\omega}^{0},\bold{e}_{0}> = 1$$

but

$$\bold{\omega}^{\alpha}\otimes\bold{e}_{\beta} = \left(\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$$

So a contraction on the 2nd expression would yield $$tr(\bold{\omega}^{\alpha}\otimes\bold{e}_{\beta}) = 4$$
Your reply has helped. I did forget that the indices are indicating which 4-component geometric object is being considered. Does the vector and 1-form have 4 components in the 1st expression, but only one in the 2nd??

"But the quantity $$\omega^{\alpha}\otimes e_{\beta}$$ is a set of 16 rank two tensors, ..."

It is a set of 16 tensors, each of rank 2! So then the contraction is done on each of these 16 tensors of rank 2. So $$\bold{e}_{0}$$ and $$\bold{\omega}^{0}$$ both have 4 components. When you take their direct product you get a 4x4 tensor. Ok, now what is the contraction on these tensors? The same as the inner product?? (I think I am beginning to see...)

Last edited: Sep 2, 2010
4. Sep 2, 2010

### phyzguy

So you have:
$$<\bold{\omega}^0,\bold{e}_0>=1$$
and:
$$\bold{\omega}^0\otimes\bold{e}_0 = \left(\begin{array}{cccc}1&0&0&0\\0&0&0&0\\0&0&0&0 \\0&0&0&0\end{array}\right)$$
So when you contract the rank two tensor you get a scalar.
$$\bold{\omega}^{\alpha}\otimes\bold{e}_{\beta}$$
Is a set of 16 such tensors.

5. Sep 2, 2010

### Living_Dog

$$\bold{\omega}^0\otimes\bold{e}_1 = \left(\begin{array}{cccc}0&1&0&0\\0&0&0&0\\0&0&0&0 \\0&0&0&0\end{array}\right)$$

$$\bold{\omega}^1\otimes\bold{e}_2 = \left(\begin{array}{cccc}0&0&0&0\\0&0&1&0\\0&0&0&0 \\0&0&0&0\end{array}\right)$$

$$\bold{\omega}^2\otimes\bold{e}_3 = \left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&0&1 \\0&0&0&0\end{array}\right)$$

etc...

So if I contract any of these 2nd rank tensors I will get '1' when the indices are along the diagonal (just as when I take the inner-product of any $$<\bold{\omega}^{\alpha},\bold{e}_{\alpha}>$$) and '0' when they aren't - like the ones I wrote above.

OHhhhh... I see now my confusion. I thought that the direct product yielded a 16 element tensor with ... "dyadic" type elements in their respective slots!!! Wow. They're not - they're those things above.

Thanks ever so much for all your help and patience with me phyzguy. May God richly bless you, in Jesus' name, amen.

-joe