Editing Tensor algebra (section)

=== Coproduct ===
The coalgebra is obtained by defining a [[coproduct]] or diagonal operator

:<math>\Delta: TV\to TV\boxtimes TV</math>

Here, <math>TV</math> is used as a short-hand for <math>T(V)</math> to avoid an explosion of parentheses. The <math>\boxtimes</math> symbol is used to denote the "external" tensor product, needed for the definition of a coalgebra.  It is being used to distinguish it from the "internal" tensor product <math>\otimes</math>, which is already being used to denote multiplication in the tensor algebra (see the section ''Multiplication'', below, for further clarification on this issue). In order to avoid confusion between these two symbols, most texts will replace <math>\otimes</math> by a plain dot, or even drop it altogether, with the understanding that it is implied from context. This then allows the <math>\otimes</math> symbol to be used in place of the <math>\boxtimes</math> symbol. This is not done below, and the two symbols are used independently and explicitly, so as to show the proper location of each.  The result is a bit more verbose, but should be easier to comprehend.

The definition of the operator <math>\Delta</math> is most easily built up in stages, first by defining it for elements <math>v\in V\subset TV</math> and then by homomorphically extending it to the whole algebra.  A suitable choice for the coproduct is then

:<math>\Delta: v \mapsto v\boxtimes 1 + 1\boxtimes v</math>
and
:<math>\Delta: 1 \mapsto 1 \boxtimes 1</math>

where <math>1\in K=T^0V\subset TV</math> is the unit of the field <math>K</math>. By linearity, one obviously has
:<math>\Delta(k)=k(1\boxtimes 1)=k\boxtimes 1=1\boxtimes k</math>

for all <math>k\in K.</math> It is straightforward to verify that this definition satisfies the axioms of a coalgebra: that is, that
:<math>(\mathrm{id}_{TV} \boxtimes \Delta) \circ \Delta = (\Delta \boxtimes \mathrm{id}_{TV}) \circ \Delta</math>

where <math>\mathrm{id}_{TV}: x\mapsto x</math> is the identity map on <math>TV</math>.  Indeed, one gets
:<math>((\mathrm{id}_{TV} \boxtimes \Delta) \circ \Delta)(v) =
v\boxtimes 1 \boxtimes 1 + 1\boxtimes v \boxtimes 1 + 1 \boxtimes 1 \boxtimes v</math>
and likewise for the other side. At this point, one could invoke a lemma, and say that <math>\Delta</math> extends trivially, by linearity, to all of <math>TV</math>, because <math>TV</math> is a [[free object]] and <math>V</math> is a [[generator (mathematics)|generator]] of the free algebra, and <math>\Delta</math> is a homomorphism. However, it is insightful to provide explicit expressions. So, for <math>v\otimes w \in T^2V</math>, one has (by definition) the homomorphism

:<math>\Delta: v\otimes w \mapsto \Delta(v)\otimes \Delta(w)</math>
Expanding, one has
:<math>\begin{align} \Delta (v\otimes w) &= (v\boxtimes 1 + 1\boxtimes v) \otimes (w\boxtimes 1 + 1\boxtimes w) \\
&= (v\otimes w) \boxtimes 1 + v\boxtimes w +  w\boxtimes v + 1 \boxtimes (v\otimes w) \end{align}</math>

In the above expansion, there is no need to ever write <math>1\otimes v</math> as this is just plain-old scalar multiplication in the algebra; that is, one trivially has that <math>1\otimes v = 1\cdot v = v.</math>

The extension above preserves the algebra grading. That is,
:<math>\Delta: T^2V \to \bigoplus_{k=0}^2 T^kV \boxtimes T^{2-k}V</math>

Continuing in this fashion, one can obtain an explicit expression for the coproduct acting on a homogenous element of order ''m'':
:<math>\begin{align}
\Delta(v_1\otimes\cdots\otimes v_m) &=
\Delta(v_1)\otimes\cdots\otimes\Delta(v_m) \\
&= \sum_{p=0}^m \left(v_1\otimes \cdots \otimes v_p\right) \;\omega
\; \left(v_{p+1}\otimes \cdots \otimes v_m\right) \\ 
&= \sum_{p=0}^m \; \sum_{\sigma\in\mathrm{Sh}(p,m-p)} \;
\left(v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(p)}\right) \boxtimes 
\left(v_{\sigma(p+1)}\otimes\dots\otimes v_{\sigma(m)}\right)
\end{align}</math>
where the <math>\omega</math> symbol, which should appear as ш, the sha, denotes the [[shuffle product]]. This is expressed in the second summation, which is taken over all [[(p,q) shuffle|(''p'', ''m'' − ''p'')-shuffles]]. The shuffle is

:<math>\begin{aligned}
\operatorname{Sh}(p,q)
= \{\sigma:\{1,\dots,p+q\}\to\{1,\dots,p+q\}\;\mid
\;&\sigma \text{ is bijective},\;\sigma(1)<\sigma(2)< \cdots < \sigma(p),\\
&\text{and }\;\sigma(p+1) <\sigma(p+2)<\cdots < \sigma(m)\}.
\end{aligned}</math>

By convention, one takes that Sh(''m,''0) and Sh(0,''m'') equals {id: {1, ..., ''m''} → {1, ..., ''m''<nowiki>}}. It is also convenient to take the pure tensor products 
</nowiki><math>v_{\sigma(1)}\otimes\dots\otimes v_{\sigma(p)}</math> and <math>v_{\sigma(p+1)}\otimes\dots\otimes v_{\sigma(m)}</math>
to equal 1 for ''p'' = 0 and ''p'' = ''m'', respectively (the empty product in <math>TV</math>). The shuffle follows directly from the first axiom of a co-algebra: the relative order of the elements <math>v_k</math> is ''preserved'' in the riffle shuffle: the riffle shuffle merely splits the ordered sequence into two ordered sequences, one on the left, and one on the right.

Equivalently,

:<math>\Delta(v_1\otimes\cdots\otimes v_n)
= \sum_{S\subseteq \{1,\dots,n\}}
\left(\prod_{k=1 \atop k \in S}^n v_k\right)
\boxtimes
\left(\prod_{k=1 \atop k \notin S}^n v_k\right)\!,</math>

where the products are in <math>TV</math>, and where the sum is over all subsets of <math>\{1,\dots,n\}</math>.

As before, the algebra grading is preserved:
:<math>\Delta: T^mV \to \bigoplus_{k=0}^m T^kV \boxtimes T^{(m-k)}V</math>