Defining Functions on Tensor Products

[WWGD]

Hi all. This question is related to my previous one on tensor products:

Is there a way of "well-defining" a function on a tensor product M(x)N (where M,N are
both R-modules) ?

This is the motivating example for my question : Say we want to define a map f: M(x)M-->M by f(m(x)m')=m+m' . But then we have that (m(x)m')=(-m(x)-m') , so that f(m(x)m')=
f(-m(x)-m')=-m-m' . But in most cases we do not have m+m' =-m-m' , so this function is not well-defined, i.e., its value depends on the choice of representative .

Clearly we must find a way of defining a function that is constant in the classes (m(x)n) , but, how do we do that? One way would be to find a bilinear map defined on MxN and determining its image on M(x)N under the map (m,n)--> (m(x)n) , since the commutative triangle guarantees that the map is well-defined. Is there some other way of doing this, i.e., of "well-defining" maps on M(x)N ?

I suspect a necessary and sufficient condition is that a candidate function satisfies the tensor
relations:

i ) ((a+a')(x)b) = (a(x)b)+(a'(x)b)

ii) (a (x) (b+b'))=(a(x)b)+ a(x)b'

iii) r(a(x)b)=(ra(x)b)= (a(x)rb)

but I can't see a good way of proving this. Anyone?

Thanks.

 

[Terandol]

The concept of a linear map on a tensor product is essentially exactly the same as a bilinear map (if the ring is commutative, this shoulde be R-balanced not bilinear but I will just use the term bilinear throughout.) It is not just, as you mentioned, that every bilinear map $\varphi: M\times N\to P$ yields an induced linear map $\tilde \varphi:M\otimes N\to P$ which agrees on simple tensors, but in fact there is a bijection between bilinear maps $\varphi:M\times N\to P$ and linear maps $\psi:M\otimes N\to P$.

The one direction is given by the universal property you mentioned and the other direction is just composing a map with the canonical inclusion of $M\times N$ into $M\otimes N$ Ie. it is defined by sending $\psi: M\otimes N\to P$ to the function $\tilde \psi:M\times N\to P$ defined by $\tilde\psi(m,n)= \psi(m\otimes n)$, which is clearly bilinear if $\psi$ is linear.

So there do not exist any linear maps on tensor products which do not come from a bilinear map on the cartesian product. In other words, bilinearity is exactly the necessary and sufficient condition required of a map defined on simple tensors.

Note that if you define a map on simple tensors, then checking that it is well-defined with respect to the equivalence relation on the tensor product is exactly the same as checking that it is bilinear (or R-balanced in the noncommutative case). So, a map defined on $M\times N\subseteq M\otimes N$ being well defined with respect to your conditions (i)-(iii) is a necessary and sufficient condition to yielding a linear map by extending to the entire tensor product. To prove this just do the trivial verification that such a map respects the relations if and only if it is bilinear.

 

[WWGD]

Thank you, nice answer. I agree that this correspondence is a bijection. I was also considering non-linear maps on the tensor product.

A follow up: I found an actual specific form for the universal bilinear map when tensoring $\mathbb R^2$ with itself, using the standard bases ${e_1, e_2}$ for both copies of $\mathbb R^2$ , and I wonder if there is a general way , given the choice of bases, to construct this universal bilinear map. In this case, the form was given by:

(1,0)(x)(1,0) <-->(1 0 0 0 )
(1,0)(x)(0,1) <-->( 0 1 0 0)
(0,1)(x)(1,0) <-->( 0 0 1 0)
(0 ,1)(x)(0,1)<-->( 0 0 0 1)

Is there a nice way for finding the universal linear map given a choice of bases?

 

[Terandol]

I assumed you meant linear maps since you were only defining your functions on simple tensors to begin with. If you care about nonlinear maps then it isn't enough to define them just on simple tensors. To use the example you gave in your original post, defining $f(m\otimes m') =m+m'$ does not even define a set function from $M\otimes M\to M$ no matter if it is bilinear or not on simple tensors when we do not require f to be linear (for example, in $M=\mathbb{R^2}$ it is impossible to write $(1,0)\otimes (0,1)+(0,1)\otimes (1,0)$ in the form $x\otimes y$ for $x,y\in \mathbb{R}^2$ so what value should f take here?)

I guess another way of saying this is that you can't just define a map on simple tensors (ie. on $m\otimes n$ for all $m\in M, n\in N$) in the nonlinear case, you have to define a function on the free abelian group with basis $M\times N$. If you do this, then of course you get a well defined map on the tensor product if and only if the map respects the conditions (i)-(iii) you gave (formally this is just an application of the universal property for quotient spaces since by definition the tensor product is a quotient space of this free abelian group.)

I'm not sure that I understand what you mean in the second part of your post. What do you mean by the universal bilinear map? The only canonical bilinear map associated to a tensor product that immediately pops into my mind is the map $\iota:M\times N\to M\otimes N$. If this is what you are talking about, then this has the very simple formula $\iota(x,y)=x\otimes y$ always but I'm not sure where the 4-component vectors came from then. From your map it looks like perhaps you mean the isomorphism $\mathbb{R}^2\otimes_\mathbb{R} \mathbb{R}^2\cong \mathbb{R}^4 ?$ If so this should be a linear map not bilinear but can be written as you've described by ordering the tensor product basis $\{ e_i\otimes e_j \}$ appropriately.

Again, what is the universal linear map you are talking about in the last line? If you start with any bilinear map then you can talk about the canonically associated linear map and this has a simple description in terms of the original bilinear map $\phi$ given by
$$\tilde \phi\left(\sum_{i=1}^n m_i\otimes n_i\right)=\sum_{i=1}^n \phi(m_i,n_i).$$

So, if you are dealing with fields (or free modules since otherwise the module does not have a basis,) you can simply compute
$$\tilde\phi(e_i\otimes e_j)=\phi(e_i,e_j)=\sum_{k,l=1}^n a^{ij}_{kl}e_{k}\otimes e_{l}$$
and then the (ij)-column of your matrix representation for the linear map will have the $n^2$ entries $a^{ij}_{kl}$. (Here I am using the slightly nontrivial fact that if $\{e_1,\cdots, e_n\}$ is a basis for $M$ then $\{ e_i\otimes e_j: 1\leq i,j,\leq n\}$ is a basis for $M\otimes M$.)

There are a few other canonical linear maps associated with a tensor product so perhaps you mean something completely different here.

 

[Terandol]

Just to clarify, when I was talking about the canonical inclusion and wrote $M\times N\subseteq M\otimes N$ this is of course wrong...The map $\iota:M\times N\to M\otimes N$ is not always injective so I shouldn't call it an inclusion (it isn't even a module homomorphism so this map isn't even in the right category) and $M\times N$ is often not contained inside the tensor product in any way. All I meant when I said this in my first post was defining a map on simple tensors rather than the whole tensor product. Sorry if this caused any confusion.

 

[WWGD]

Excellent, thanks for everything, Terandol. Tensor products have been my master of puppets for a while. I know it is a matter of sitting down and dotting t's and crossing i's, which I am doing now.

 

[WWGD]

Sorry, Terandol, one last one , hopefully to put a nail on the coffin;I guess the result we are using to justify the existence of the tensor product is just the universal property of the tensor product :

Let $H \triangleleft G$ (i.e., H is normal in G; normality follows here from the fact that the free module on $V\times W$ is Abelian), and let $\phi: G \rightarrow K$ be a group homomorphism such that $H \subset \ker \phi$ . The Universal Property of the Quotient says that there is a unique homomorphism $\tilde\phi: G/H \rightarrow K$ such that $\tilde\phi\cdot \pi = \phi$ , where $\pi: G \to G/H$ is the quotient map .

 

[Terandol]

I think you have the general idea correct but your logic seems slightly confused here. The fact that the tensor product exists in the category of R-modules is not a result of the universal property of the tensor product. In general whenever you define things in a category in terms of a universal property, you are automatically guaranteed that if the thing exists, then it is unique however you are not guaranteed existence.

So if you define the module tensor product using the universal property, the first thing you have to do is prove it actually exists. To do this, you actually have to construct it which is done using a free module and then taking the quotient as you already know. Then you have to show that the thing you have constructed does indeed satisfy the universal product. Here is where the universal property of quotients of free modules comes up (this is a completely different universal property than the tensor product one though.)

Your last paragraph gives part of the proof that the quotient of our free module does satisfy the required universal property of tensor products since $H\subseteq \ker \phi$ where H is the submodule generated by the tensor product relations is equivalent to saying the map is bilinear. However, you are also using the universal property of free modules to extend the map from $M\times N$ to the free module $F^{M\times N}$ to begin with and then you apply the universal property of quotients to get a map on the tensor product.

So the existence of the tensor product relies on this explicit construction and the easiest way to prove this construction satisfies the universal property you want it to satisfy uses both the universal property of free modules and the universal property of quotients to construct a unique linear map from the given bilinear one.