Tensor Product of Algebras: Multiplication Definition

I've managed to get a weak grip on what the tensor product of two vector spaces is. I'm now trying to understand the tensor product of two algebras.

I understand that we define $(v_1\otimes w_1)(v_2\otimes w_2):=(v_1v_2)\otimes (w_1w_2)$. In order for this to be bilinear we must have relations like $x(y+z)=xy+xz$ (x,y,z are tensors here). I can't show that this is true: you can't in general simplify two tensors $y$ and $z$ into a new tensor of the form $v\otimes w$ (i.e the sum won't necessarily be a pure tensor). But then the multiplication isn't defined for a sum of pure tensors, only for pure tensors. What can I do here? Thanks for any replies!

The equation $(v_1\otimes w_1)(v_2\otimes w_2)=(v_1v_1\otimes w_1w_2)$ tells us how to multiply pure tensors but does not immediately say how to multiply sums of pure tensors. But there's only one possible way to extend the multiplication operation to sums of pure tensors in a way that the distributive property holds, and that is through the distributive property itself: when multiplying sums of pure tensors, first distribute and then use the given rule. It is very often in situations like these that we only define multiplication for a set of things and then "extend linearly" to all else via distributivity.

• Oh ok, so isn't that basically including distributivity in the definition of the multiplication? – James Machin May 15 '14 at 19:25
• More or less. The rule still needs to be checked whenever sums of pure tensors actually do simplify to pure tensors (i.e. multiplications involving expressions like $(v_1+v_2)\otimes w$). – blue May 15 '14 at 19:27
• Ah ok, well thanks a lot! – James Machin May 15 '14 at 19:33

A linear map from a tensor product $A \otimes B$ is defined by first constructing a bilinear map from $A \times B$ and checking that scalars commute across the product. Specifically, a bilinear map $\varphi: A \times B \to C$ that also satisfies $$\varphi(at, b) = \varphi(a, tb)$$ where $a \in A$, $b \in B$, and $t$ is a scalar descends to a map $$\overline{\varphi}: A \otimes B \to C,$$ defined on pure tensors by $\overline{\varphi}(a \otimes b) = \varphi(a, b)$. The map is then extended by linearity, as pure tensors form a linear spanning set for $A \otimes B$.

You can think of multiplication as a bilinear map $$\left( A \otimes B \right) \times \left( A \otimes B \right) \to A \otimes B,$$ or, equivalently, as a linear map $$\left( A \otimes B \right) \otimes \left( A \otimes B \right) \to A \otimes B.$$ Since tensor products are associative, this is a map from the four-fold tensor product, which is most easily constructed as the composition: $$A \otimes B \otimes A \otimes B \xrightarrow{~~\tau~~} A \otimes A \otimes B \otimes B \xrightarrow{m_A \otimes m_B} A \otimes B,$$ where $\tau$ is the permutation map on the second and third tensor factors and $m_A$ and $m_B$ are the multiplication maps on the respective algebras.

• Thankyou very much! Does this mean that multiplication on a algebra $A$ is a linear map from $A\otimes A\rightarrow A$? It's really helping my understanding, but how does this explain the distributivity of multiplication? – James Machin May 15 '14 at 19:45
• That's exactly right. The bilinearity of multiplication $A \times A \xrightarrow{m_A} A$ is equivalent to linearity of the map $A \otimes A \xrightarrow{\overline{m}_A} A$. – Sammy Black May 15 '14 at 19:47
• By abuse, these two maps are often denoted by the same symbol. (I seem to have committed this abuse in my answer, where the multiplication maps ought to have been wearing overbars). – Sammy Black May 15 '14 at 19:48