what the fuck are tensors more importantly what are covariant,

A tensor is set of quantities that obey certain transformation laws relating the bases in one generalized coordinate system to those of another and involving partial derivative sums. vectors are simple tensors.

Because of this you can think of tensors as a generalized form of vectors.

A tensor is contravariant when it transforms according to the following equation

[math]A^i=\frac{\partial y^i}{\partial x^j}B^j[/math]

Note that i and j are indcies not exponents. A tensor is covariant if it transforms according to the following equation

[math]A_i=\frac{\partial x^j}{\partial y^i}B_j[/math]

A mixed tensor is a tensor that contains both contravariant indices and covariant indices.

A tensor should be thought of as a multi linear operator that transforms (between different coordinate systems) according to certain rules (to be detailed below).

A rank 2 tensor takes two vectors as inputs. This gives it two indicies, which can be covariant, contravariant, or one of each (mixed).

Covariant index transforms by the inverse jacobian and contravariant transforms by the jacobian.

A covariant index takes a contravariant vector as its argument, and vice versa. This is because the contravariant transforms by the jacobian and the covariant transforms as the jacobian, so when they are multiplied together their transformation matrices cancel out and the result is coordinate independent.

File: IMG-20150918-WA0005.jpg (63KB, 480x556px) Image search: [Google]

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>tfw /sci/ has actually good answers

>>7924729
I forgot to mention this but those equations are written using Einstein summation convention

>>7924729
>>7924730
You guys both suck. Let me check if the LaTeX works as expected here: [math] x^2 [/math}

>>7924742
I suppose that ought to be [math] x^2 [/math]

>>7924742
Lets see you answer OP's question

>>7924698
do you do math or physics (don't say both, then it's physics) ?

>>7924729
>>7924730

thanks guys, mind sharing an example if that's not too much to ask?

>>7924754
more on the physics side tho you could say both

>>7924698
Ok, so forget what the two first plebs told you. Let [math]V[\math] and [math]W[\math] be vector spaces over a field [math]k[\math]. Define [math]V \otimes_k W[\math] by the following universal property: any bilinear map [math]V \times W \to W'[\math] factors uniquely as a linear map through the tensor product. Prove as an exercise that this is unique up to isomorphism. Now, mixed tensors are of the form [math]V^{\otimes r} \otimes_k \text{Hom}(V, k)^{\otimes s}[\math]. These things may be identified with multilinear maps from the regular product (direct sum) to the ground field, as is often the definition taught to physics shitfaces.

>>7924765
Too autistic to remember how to TeX here.

Ok, so forget what the two first plebs told you. Let [math]V[/math] and [math]W[/math] be vector spaces over a field [math]k[/math]. Define [math]V \otimes_k W[/math] by the following universal property: any bilinear map [math]V \times W \to W'[/math] factors uniquely as a linear map through the tensor product. Prove as an exercise that this is unique up to isomorphism. Now, mixed tensors are of the form [math]V^{\otimes r} \otimes_k \text{Hom}(V, k)^{\otimes s}[/math]. These things may be identified with multilinear maps from the regular product (direct sum) to the ground field, as is often the definition taught to physics shitfaces.

File: Another_universal_tensor_prod.svg.png (4KB, 248x142px) Image search: [Google]

4KB, 248x142px

>>7924698
A tensor is element of a representation of the space defined in the category of multilinear maps via the universal property in pic related.

For your purposes V and W are some self-dual vector spaces. You assign a index for each original vector space and for each duality (for a tensor with n indices, you have up to n different such maps, although depending on the scenario you might e.g. use the same matrix [math]g_{\mu\nu}[/math] for each) you get a map to a similar tensor space, so [math]2^n[/math] in total, e.g. [math]4=2^2[/math] for [math]n=2[/math].

[math] T_{ij}[/math]
[math] T^i_j = \sum_k \ g^{ik} \ T_{kj} [/math]
[math] T_i^j = \sum_r \ g^{jr} \ T_{ir} [/math]
[math] T_i^j = \sum_k \sum_r g^{ik} \ g^{jr} \ T_{rk} [/math]

From the self-duality, it follows that co- and contravariance is basically just indexing (meme book keeping).

Basis change within the spaces force some transformation laws on the tensors also, though it's a little bit cryptic to take that as the definition tbqh. And I say that as a physicist.

>>7924770
>Hom(V,k)
That must be the homs in the cateogry of notation abuse

>>7924764
Take some matrix A. Then fix a basis in some vector space. You should know that now you can interpret A as a linear mapping (its columns are coordinates of images of basis vectors) or as a bilinear form (its elements are the values of the form on basis vectors). And as a shitload of other things, but let's stick to these two. The point is, if you change the basis, A will change differently if you think of it as a linear map or as a bilinear form. The reason is that a linear map and a bilinear form are tensors of a different type. It's kinda hard to grasp the actual definition but the purpose of tensors is to generalize this concept of "change of variables" and "basis independence". This is of course crucial in the theory of manifolds as you don't have any cannonical coordinates (as opposed to Rn) and you have to define things so they won't depend on them. so if this is your first encounter with tensors, the definition "it's a bunch of numbers which transforms under certain law" is really the best one for you I think

>>7924698
what is contraction?

>>7924841
Contraction is basically an operation performed on tensors that produces a new tensor of order [math]r-2[/math]

If contraction and multiplication occur together then the operation is called inner multiplication

I'm too lazy to work out an example in TeX

>>7924742
This is the greatest post this week. Thanks for the laugh

>>7924698
A tensor is an object that transforms as a tensor.

>>7924770
Is there a point to this approach, other than the "look how cool I am!" factor. You can define a tensor the old fashioned way and prove it has certain properties... but no, that's so lame. Better to write down the properties first and show there exists a unique structure that has those properties.

But what's the difference really?

>>7925110
It applies much more generally, which turns out to be what you want in algebra and geometry (aside from smooth manifolds).

File: hNu0Kivt3cs.jpg (60KB, 520x390px) Image search: [Google]

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>>7924839
every book or course about tensors should start with what's in this post. Thanks anon

>>7925110
The first definitions in this thread define a tensor more directly, that guy you responded to was rather setting up the spaces the tensor is an element of.
Generally though, pointing out where there are vector spaces and dual spaces involved, i.e. common algebraic structures with many theorems available about them, should be done if possible, at least if you're doing math.
It's quite possible that OP is going though his undergrad study and is happy once he got that stuff over with, going into some engineering job and never see math again. If he only has to deal with it for a year, a definition where everything is some matrix and indices are seemingly arbitrarily moved up and down might work.

The "unique" aspect is part of the (universal) property, you don't have to show that there is a -unique- structure with that property, you might just show that there is one and then each isomorphism in the category gives another. Although that might be pedantic language. (What's unique is the morphism h-bar in the pic in >>7924815.)

A difference is that the universal property is all that's ever used in algebra. E.g. while you might define an ordered pair [math](a,b)\in A\times B[/math] as a certain set in set theory [math](a,b):=\{\{a\},\{a,b\}\}[/math], model depend facts like [math]\{\{a\}\}=(a,a)[/math] are provable but never used. What's used is the defining property
[math](a,b)=(a_2,b_2) \leftrightarrow a=a_2 \ \land \ b=b_2 [/math]

Working with the tensor space in terms of its defining property in a category with multilinear maps makes working with it clear - you don't ever need to think about the ridiculously huge machinery of equivalence classes involved in defining [math]\otimes[/math] in terms of those vector spaces (involving sets of vectors).
If you're just doing proof sketches, working with the later definition directly might still work, although I expect the universal propery proofs to be much simpler anyway.

>>7925297
Thanks anon.

>>7925557
I can't really claim I had convinced myself when posting the answer, though.
There is maybe too little time in the curricula to bring the coordinate free'ish definitions.

>>7925568
>free'ish
fag'ish word fgt pls

Feynmann seems to have a good chapter on it:

http://www.feynmanlectures.caltech.edu/II_31.html

Though this is where I'm learning what a tensor is, so I have no clue whether it explains it properly or not, as it's my only real exposure.

Here you go OP, watch this
https://www.youtube.com/watch?v=mbv3T15nWq0