Editing Tensor algebra (section)

== Bialgebra==
A [[bialgebra]] defines both multiplication, and comultiplication, and requires them to be compatible.

=== Multiplication ===
Multiplication is given by an operator
:<math>\nabla: TV\boxtimes TV\to TV</math>
which, in this case, was already given as the "internal" tensor product. That is,
:<math>\nabla: x\boxtimes y\mapsto x \otimes y</math>
That is, <math>\nabla(x\boxtimes y) = x \otimes y.</math> The above should make it clear why the <math>\boxtimes</math> symbol needs to be used: the <math>\otimes</math> was actually one and the same thing as <math>\nabla</math>; and notational sloppiness here would lead to utter chaos.  To strengthen this: the tensor product <math>\otimes</math> of the tensor algebra corresponds to the multiplication <math>\nabla</math> used in the definition of an algebra, whereas the tensor product <math>\boxtimes</math> is the one required in the definition of comultiplication in a coalgebra.  These two tensor products are ''not'' the same thing!

=== Unit ===
The unit for the algebra
:<math>\eta: K\to TV</math>
is just the embedding, so that
:<math>\eta: k\mapsto k</math>
That the unit is compatible with the tensor product <math>\otimes</math> is "trivial": it is just part of the standard definition of the tensor product of vector spaces.  That is, <math>k\otimes x = kx</math> for field element ''k'' and any <math>x\in TV.</math>  More verbosely, the axioms for an [[associative algebra]] require the two homomorphisms (or commuting diagrams):
:<math>\nabla\circ(\eta \boxtimes\mathrm{id}_{TV}) = \eta\otimes \mathrm{id}_{TV} = \eta\cdot \mathrm{id}_{TV}</math>
on <math>K\boxtimes TV</math>, and that symmetrically, on <math>TV\boxtimes K</math>, that
:<math>\nabla\circ(\mathrm{id}_{TV}\boxtimes\eta) = \mathrm{id}_{TV}\otimes\eta = \mathrm{id}_{TV}\cdot\eta</math>
where the right-hand side of these equations should be understood as the scalar product.

=== Compatibility ===
The unit and counit, and multiplication and comultiplication, all have to satisfy compatibility conditions. It is straightforward to see that 
:<math>\epsilon \circ \eta = \mathrm{id}_K.</math>
Similarly, the unit is compatible with comultiplication:
:<math>\Delta \circ \eta = \eta \boxtimes \eta \cong \eta</math>
The above requires the use of the isomorphism <math>K\boxtimes K \cong K</math> in order to work; without this, one loses linearity. Component-wise,
:<math>(\Delta \circ \eta)(k) = \Delta(k) = k(1 \boxtimes 1) \cong k </math>
with the right-hand side making use of the isomorphism.

Multiplication and the counit are compatible:
:<math>(\epsilon \circ \nabla)(x\boxtimes y) = \epsilon(x\otimes y) = 0</math>
whenever ''x'' or ''y'' are not elements of <math>K</math>, and otherwise, one has scalar multiplication on the field: <math>k_1\otimes k_2=k_1 k_2.</math>  The most difficult to verify is the compatibility of multiplication and comultiplication:
:<math>\Delta \circ\nabla = (\nabla \boxtimes \nabla) 
\circ (\mathrm{id} \boxtimes \tau \boxtimes \mathrm{id}) 
\circ (\Delta \boxtimes \Delta)</math>
where <math>\tau(x\boxtimes y)= y \boxtimes x</math> exchanges elements. The compatibility condition only needs to be verified on <math>V\subset TV</math>; the full compatibility follows as a homomorphic extension to all of <math>TV.</math> The verification is verbose but straightforward; it is not given here, except for the final result:
:<math>(\Delta \circ\nabla)(v\boxtimes w) = \Delta(v\otimes w)</math>
For <math>v,w\in V,</math> an explicit expression for this was given in the coalgebra section, above.