Prove a tensor product

The tensor product of two tensors is a tensor of higher order. What do you mean by "forms the basis"?

 

I find that an odd saying. Could you please reference several physics texts which say this so that the likelyhood that I'll have one of them will be good? Thanks

Best wishes

Pete

 

That one I have. What page should I turn to?

Pete

 

If you look at page 71 you'll see that Schutz says quite explicitly

This contradicts what you've said above. In fact, Schutz goes to great lengths to explain how and why the most general (0,2) tensor must be a sum of tensor product terms. What, specifically, is the difficulty that you're having?

 

I wouldn't worry if you're in high school and finding tensor analysis quite difficult!

He introduces [tex]x^{\alpha}_{, \beta}\equiv \delta^{\alpha}_{\beta}[/tex] [sorry, I dont know how to offset the indices like in the text]. Note he is using , to denote partial derivative, so the LHS is [tex]\frac{\partial x^{\alpha}}{\partial x^{\beta}}[/tex] Do you know what the kronecker delta is?

The conclusion comes from looking at (3.12). Do you understand this equation?

alpha and beta can take the values {0,1,2,3}, and so the tensor [itex]f_{\alpha \beta}[/itex] has components f_00, f_01, f_02, ..., f_33. There are 16 in total.

 

The kronecker delta is defined as (im switching to latin indices, as they're easier to type!) [tex]\delta^a_b=\left\{\begin{array}{cc}1,&\mbox{ if } a=b \\1, & \mbox{ if } a\neq b \end{array}\right\}[/tex], so, (3.12) simply says that a basis oneform [itex]\tilde{\omega}^a[/itex] acting on a basis vector [itex]\vec{e}_b[/itex] is 0 if [itex]a\neq b[/itex] and is 1 if a=b. Comparing this with [itex]x^{\alpha}_{, \beta}\equiv \delta^{\alpha}_{\beta}[/itex] gives us the result that the basis one form is [itex]\tilde{d}x^a[/itex].

Now, to explain the 16 components. Let f be the arbitrary tensor. The components [itex]f_ab[/itex] of this tensor are obtained by contracting f with the basis vector e_a in the first slot, and e_b in the second slot. We have 4 choices for each basis vector, and so there are 4x4=16 components (since for each choice of basis vector in slot one, we have 4 choices for the basis vector in slot 2).

 

It might be helpful to consider the following when trying to understand the Kronecker delta. Suppose that you have some system of coordinates [itex]x^a[/itex], where [itex]a=0,1,2,\ldots,m[/itex]. Now consider the partial derivative

[tex]\frac{\partial x^a}{\partial x^b} = x^a_{,b}[/itex]

Schutz tells you that [itex]x^a_{,b}=\delta^a_{\phantom{a}b}[/itex]. To see why this is so, consider a simple example where [itex]a=0,1[/itex]. Then [itex]x^a_{\phantom{a},b}[/itex] can be represented as a matrix:

[tex]x^a_{\phantom{a},b} =
\left(
\begin{array}{cc}
\frac{\partial x^0}{\partial x^0} & \frac{\partial x^0}{\partial x^1} \\
\frac{\partial x^1}{\partial x^0} & \frac{\partial x^1}{\partial x^1}
\end{array}
\right) =
\left(
\begin{array}{cc}
1 & 0 \\ 0 & 1
\end{array}
\right) = \delta^{a}_{\phantom{a}b}
[/tex]

The reason that, for example, [itex]\partial x^0/\partial x^1 = 0[/itex] while [itex]\partial x^0/\partial x^0=1[/itex] should be obvious. If it isn't, note that our coordinates are supposed to be independent, so if you differentiate [itex]x^0[/itex] with respect to [itex]x^1[/itex] you will get zero, while differentiating [itex]x^0[/itex] with respect to [itex]x^0[/itex] will give you 1. So it should be easy to see why the definition of the Kronecker delta allows you to write this. Extending things to a scenario where you have [itex]m[/itex] coordinates is then trivial:

[tex]
x^a_{\phantom{a},b}
=
\left(
\begin{array}{cccc}
\frac{\partial x^0}{\partial x^0} & \frac{\partial x^0}{\partial x^1} &
\cdots & \frac{\partial x^0}{\partial x^m} \\
\frac{\partial x^1}{\partial x^0} & \frac{\partial x^1}{\partial x^1} &
\cdots & \frac{\partial x^1}{\partial x^m} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{\partial x^m}{\partial x^0} & \frac{\partial x^m}{\partial x^1} &
\cdots & \frac{\partial x^m}{\partial x^m}
\end{array}
\right)
=
\left(
\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{array}
\right)
= \delta^{a}_{\phantom{a}b}.
[/tex]

 

Its probably one of the most difficult branches of math that there are. Graduate students have trouble with that math too. So if you're having trouble then you're normal.

Pete

 

high schoolers? uh, no it is not normal for high schoolers to find it difficult, because most high schoolers are prudent enough to leave the topic for 4-5 years later. do you understand trig well? plane geometry? solid geometry? logic? algebra of polynomials? probability? matrices? calculus of one and several variables? topology?

if not, my suggestion is leave the tensors alone.

 

as to the original question, why are monomials a basis for all polynomials? (these are products of coordinates.) similarly, products of linear forms give basis for all multilinear functions of any degree.