Tensor product

to get a handle on tensor products, maybe you should try to figure out a few concrete products.

for example, what is Z/n * Z/m for m and n coprime / not coprime
(both are Z modules, m and n natural numbers)

or, what is Z[x] * Z[y] (both Z modules, x and y variables).

i know the answer is supposed to be zero...
so...

the simplest way to construct the tensor product V X W is to consider the group generated by the elements of the form v X w, the tensor products of elements of the two spaces. This is fairly simple to handle for vector spaces since you can just consider the basis vectors and the tensor product has very simple properties, just satisfying bilinearity of elements of the spaces. look at chapter 7 of evan chen's napkin

Do you know it because you were told so, or do you know it because you can FEEL the Z modules inside of you and understand them on a deep, emotional and personal level?

this is just a bilinear map [math] \* : \mathbb{Z} \setminus n \times \mathbb{Z} \setminus m \longrightarrow \mathbb{Z} \setminus n m [/math]

this is the best way to think of it. the technical def is that it is the "universal bilinear map out of MxN"

but you can think of it as formally building sums of formal products mxn which have to meet a "distributive law" in that ((a+b)xc)=(axc)+(bxc). It is in a sense the "best" way to endow MxN with a module structure given the "formal" multiplication.

this guy is implicitly doing something called the "free construction"

effectively, form formal sums of formal products mxn and take the quotient by the bilinear relations

i.e. demand that ((a+b)xc)=axc+bxc etc after building the formal linear combinations of formal products

Undergrad here. I see this is all about linear algebra. I am only studying inner product spaces though. When does all of this come into play?

they help generalize linear maps into something called a "multilinear" map which pretty much means "linear in each coordinate". precisely, you have some collection Vi, W of k-spaces, and a function f from the product of the Vi to W. The function is multilinear if for each space Vi and for any fixed choice of argument for each j≠i, the induced function from Vi to W is linear (varying only the ith argument of f). Every multilinear map is in correspondence with a linear map from the tensor product of the Vi to W.

In mathematical physics, I think there are some applications where you are interested in information recorded as multilinear maps from copies of the tangent space of a manifold (loosely, a vector space of all directional derivatives at a point) to another space. e.g. information about local "area" might be encoded in a multilinear map from a square of a tangent space to the space of scalars.

ps - inner products are special cases - they are bi-linear maps from VxV*-->k

the former

The tensor product of two vector spaces is defined, up to isomorphy, by the universal property in your picture. Explicitely, you can take a basis [math]\{v_1,\ldots,v_m\}[/math] of [math]M[/math] and a basis [math]\{w_1,\ldots,w_n\}[/math] of [math]N[/math], and consider the quotiont vector space [math]\mathrm{span}_{R}\{(v_i,w_j)\ |\ 1\leq i\leq m,\ 1\leq j\leq n\}/_{\sim}[/math], where [math](v,w)\sim(rv,r^{-1}w)\ \forall r\inR\setminus\{0\}[/math]. Then you show that said quotient vector space fulfils the universal property of the tensor product and, hence, is isomorphic to [math]M\otimes_R N[/math].

Sounds fun. We are only touching up to bilinear and quadratic forms this semester though.

Scalars [math] \in \mathbb{C} [/math] are tensors of order 0.
Vectors [math] \in \mathbb{C}^{n} [/math] are tensors of order 1.
Matrices [math] \in \mathbb{C}^{n} \otimes \mathbb{C}^{n} [/math] are tensors of order 2.

Let [math] \mathbf{x} = \begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{pmatrix} \& \mathbf{y} = \begin{pmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{pmatrix} [/math]
Then [math] \mathbf{x} \otimes \mathbf{y}[/math] is the matrix
[math] \mathbf{y} = \begin{pmatrix}
x_{1}y_{1} & x_{1}y_{2} & \cdots & x_{1}y_{n} \\
x_{2}y_{1} & x_{2}y_{2} & \cdots & x_{2}y_{n} \\
\vdots & \vdots & \ddots & \vdots \\
x_{m}y_{1} & x_{m}y_{2} & \cdots & x_{m}y_{n}
\end{pmatrix} [/math]

In Programming languages (such as C)
Tensor == Arrays
>
float scalar;
/* a scalar is a tensor (array) of order 0*/
float array[ i ];
/* a vector is tensor (array) of order 1*/
float array[ i ][ j ];
/* a matrix is tensor (array) of order 2*/
float array[ i ][ j ][ k ];
/* is an example of tensor (array) of order 3*/
float array[ i ][ j ][ k ][ l ];
/* is an example of tensor (array) of order 4*/
float array[ i ][ j ][ k ][ l ][ m ];
/* is an example of tensor (array) of order 5*/
float array[i_1][i_2] ... [i_n];
/* is an example of tensor (array) of order n*/

kinda meaningless as tensors are only tensors by how they act when various operations are performed on them, and these are just lists, no operations. They could just as easily be pseudotensors

Here is how the tensor product is defined in Lee Smooth Manifolds. It's really not that sophisticated.

Fugg

If you took calc 3 and wondered why there was a "positive" and a "negative" way to traverse a closed curve, it all has to do with this kind of stuff.

are you talking about pull-backs and push-forwards? or maps lets say [math] f: T_p M \longrightarrow T_q N[/math]?

ps so the inner product is a (1-1)-tensor?

yes an inner product is a (1,1) tensor.

I was not talking about either pbs or pfs. A "tensor" in the sense of "tensor field" at a point of a manfold (or even a locally ringed space, anything where you have continuously varying tangent spaces) is an (r,s) tensor over the vector field TpM. A tensor field is a selection of tensor at each point whose coordinate functions are continuous.

I'm way big into deep learning, TensorFlow is my bestie

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so you are talking about a tensor field lets say t defined as: [math] t :\Gamma (T^{\ast} M) \times \dots \times \Gamma (T^{\ast} M) \times \Gamma(TM) \times \dots \times \Gamma(TM) \longrightarrow C^{\infty}(M) [/math]

>saying tensor instead of matrice

do you want to appear 'smart' by any chance?

It's the biggest linear object, as far as linear maps into the Cartesian products go.

The old school definition is shit and the categorical definition lacks in that there's no formal definition of free functors, one that wouldn't presuppose a particular category of sets.

wtf i didnt write anything like that

Let T(r,s)M be the disjoint union of all tensors of type (r,s) on TpM for all p in M. A tensor field is a function t: M->T(r,s)M such that t(p) is a tensor of type (r,s) over TpM (I'm pretty sure you can also turn T(r,s) into a manifold with a mapping e over M taking each tensor to the point whose tangent space it lives on, and ask that t is a "section" of e). Say it is smooth if at each point, the coordinates of t(p) are smooth functions.