The h Bar

3:01 PM

give me a sec when I write up the caption

ok this is really bugging me since the day I learnt GR (and Suits (is that how his name is spelt?) and MTW etc. don't approach tensors that way)

I understand that if indices pair up it is implied summation

But how exactly do the basis components get removed so that you end up with a tensor of less rank when you do a tensor contraction?

So I have no problem on how to manipulate components, but I am confused on how exactly the rank get reduced

(obviously my calculation above is all wrong since I have odd number of indices appearing in some terms)

How to tensor contract the basis of two tensors?

Slereah

The dual basis has the property that $\omega_a (v^b) = \delta_a^b$

Secret

simplified version

Ok, that I understand, but what about the $\omega_a(\omega_j)$ and $v^a(v^i)$ terms, are they just ordinary inner products?

0celo7

Perhaps, depends how you define things I guess.

Imo T(U) doesn't make much sense

Slereah

You probably shouldn't get those :p

0celo7

3:16 PM

T is a map on a vector space of which U is not a member

Secret

now I am really confused (got hit in the head of what makes me felt so confused about GR)

So given the above simplified version, how does one expand $T \cdot U$ so that the basis and dual basis elements are in the right place (and become $\delta_i^j$)?

as in, what is the correct rules to pair up the basis elements when you expand $T\cdot U$?

0celo7

T\cdot U does not make sense

So we can't help you

It's like asking for help understanding 2\otimes 3

It does not make sense

Better analogy: you say f(x)=x^2 and you want help calculating f(sheep)

Secret

0celo7

Ok, that makes sense

That's not T(U)

Slereah

Basically what you are doing is some currying

You are using an incomplete function to generate another function

0celo7

3:25 PM

I still don't understand the issue.

You're just performing a contraction on T\otimes U

Secret

@ocelo7 From the compoennt only expression above, I see what happened to the components, but I have no idea how that reduces the rank

Because if it reduces the rank, it means some basis elements in that long sequence of tensor product has somehow gone, but how?

@Slereah currying?

0celo7

Uh the contraction is defined such that a pair of basis elements turn into a delta

And then you get the sum

Slereah

If you want, I have a fancy math book on tensors and shit

I can look up the PROPER WAY

Secret

I STRONGLY need to have a read on that, this underlies EVERY SINGLE CONFUSION I have when I learn GR back then

what's the name of the book and where I can find it?

Slereah

"Ricci calculus", by J.A. Schouten

If you want to have tensors all the way down, that's the book

It even contains words I've never seen anywhere else

Like "transvection"

0celo7

3:31 PM

"Proper way"

What is so hard about this

Slereah

Well in GR usually it's done with tensor components

Not tensors directly

0celo7

You can define it without components by taking the trace of the map on whatever slots of the tensor

Something like that

It's not illuminating

ACuriousMind

@Secret The basis elements are applied to each other. Given $T = {T^{ab}}_c \omega_a\otimes\omega_b\otimes v^c$, the contraction ${T^{ab}}_b$ is obtained by doing $\omega_b(v^c) = \delta_b^c$. Contraction of the $i$-th index with the $j$-th index just means applying the $i$-basic tensor in the expansion to the $j$-basic tensor in the expansion.

Secret

@ACuriousMind

ACuriousMind

@Secret No

Secret

3:42 PM

confused

ACuriousMind

The contraction ${T^{ab}}_c U^{cd}$ is obtained by first calculating the tensor $W := T \otimes U$ (which has components ${{W^{ab}}_c}^{de}$ and is a rank-5 tensor) and then contracting its third and fourth slot.

Idk what you did there, but it doesn't make any sense to write two tensors in full component notation beside each other without a $\otimes$ between them as in your first line after "written in full is".

Secret

@ACuriousMind You mean like this?

line 1-2 make the tensor $T \otimes U$
line 3 starts the contraction
line 4 apply the duality of basis
line 5 apply the rules of kronecker delta
line 6 move the scalar compoennt to the front
line 7 write it as compoennts again?

Slereah

No need to use a tensor product for the components

They are just real numbers

ACuriousMind

@Secret Almost. Why is there a tensor sign where $U$ stands alone? The components are not vectors/tensors, they are just numbers! From the second to third line, one more of the tensor signs should vanish.

NeuroFuzzy

4:00 PM

@Secret I was in your boat (I studied through "tensor calculus" by Kay and found it worthless) and got the book Manifolds Tensors and Forms by Renteln

Secret

@ACuriousMind but didn't you mentioned earlier that " The tensor product is not a restriction of a Cartesian product (it is larger than the Cartesian product!), and the Cartesian product of a space of scalars and a space of vectors is not scalar multipliciation (it's a vector space of one dimension more)"

thus $a \otimes T \neq aT$

Are the components just numbers in that they are not even scalars?

ACuriousMind

@Secret $aT$ is not a Cartesian product, it is scalar multiplication.

Secret

@NeuroFuzzy This is why I failed that general relativity course

@NeuroFuzzy no matter how much I like it at that time, I faield to understand how a tensor reduct its rank in dimension

ACuriousMind

You can't sensibly write $a \otimes T$ if $a$ is a scalar and $T$ a vector/tensor because this operation is not defined on the space of tensors. It just doesn't make any sense.

But the only meaningful way would be to just have $a\otimes T = aT$.

The Cartesian product never enters into this

The tensor product of a vector space with a one-dimensional vector space (which the scalars are) just gives the original vector space (precisely through the equality $a\otimes T = aT$).

NeuroFuzzy

@Secret so, a vector is something tangent to a curve (there are several choices of definition to turn it into more formal language), a covector is an element of the dual space if vectors, and the tensor product builds tensors from that. All of the transformation rules follow and you learn when to use components and when not to. Eg, $ y^i\frac {d}{dx^i}$ is a vector and nothing happens to it under a change of basis, but the components do change.

Ch 2 of renteln covers tensor products, ch 3 covers smooth manifolds (and tensors). So if you can get your hands on that book you coukd just study those.

Secret

4:09 PM

@ACuriousMind Like this?

ACuriousMind

@Secret Almost. Why is the $TU$ still tensored with the basis vectors? Don't you see that that is inconsistent with how you wrote the components at the start (without tensor sign)?

Secret

Sorry that's a typo in $\otimes$ from 2nd last line to final line

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