Tensor algebra

The coalgebra is obtained by defining a coproduct or diagonal operator

${\displaystyle \Delta :TV\to TV\boxtimes TV}$

Here, ${\displaystyle TV}$ is used as a short-hand for ${\displaystyle T(V)}$ to avoid an explosion of parentheses. The ${\displaystyle \boxtimes }$ 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 ${\displaystyle \otimes }$, which is already "taken" and 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 ${\displaystyle \otimes }$ by a plain dot, or even drop it altogether, with the understanding that it is implied from context. This then allows the ${\displaystyle \otimes }$ symbol to be used in place of the ${\displaystyle \boxtimes }$ 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 ${\displaystyle \Delta }$ is most easily built up in stages, first by defining it for elements ${\displaystyle v\in V\subset TV}$ and then by homomorphically extending it to the whole algebra. A suitable choice for the coproduct is then

${\displaystyle \Delta :v\mapsto v\boxtimes 1+1\boxtimes v}$

and

${\displaystyle \Delta :1\mapsto 1\boxtimes 1}$

where ${\displaystyle 1\in K=T^{0}V\subset TV}$ is the unit of the field ${\displaystyle K}$. By linearity, one obviously has

${\displaystyle \Delta (k)=k(1\boxtimes 1)=k\boxtimes 1=1\boxtimes k}$

for all ${\displaystyle k\in K.}$ It is straight-forward to verify that this definition satisfies the axioms of a coalgebra: that is, that

${\displaystyle (\mathrm {id} _{TV}\boxtimes \Delta )\circ \Delta =(\Delta \boxtimes \mathrm {id} _{TV})\circ \Delta }$

where ${\displaystyle \mathrm {id} _{TV}:x\mapsto x}$ is the identity map on ${\displaystyle TV}$. Indeed, one gets

${\displaystyle ((\mathrm {id} _{TV}\boxtimes \Delta )\circ \Delta )(v)=v\boxtimes 1\boxtimes 1+1\boxtimes v\boxtimes 1+1\boxtimes 1\boxtimes v}$

and likewise for the other side. At this point, one could invoke a lemma, and say that ${\displaystyle \Delta }$ extends trivially, by linearity, to all of ${\displaystyle TV}$, because ${\displaystyle TV}$ is a free object and ${\displaystyle V}$ is a generator of the free algebra, and ${\displaystyle \Delta }$ is a homomorphism. However, it is insightful to provide explicit expressions. So, for ${\displaystyle v\otimes w\in T^{2}V}$, one has (by definition) the homomorphism

${\displaystyle \Delta :v\otimes w\mapsto \Delta (v)\otimes \Delta (w)}$

Expanding, one has

${\displaystyle {\begin{aligned}\Delta (v\otimes w)&=(v\boxtimes 1+1\boxtimes v)\otimes (w\boxtimes 1+1\boxtimes w)\\&=(v\otimes w)\boxtimes 1+v\boxtimes w+w\boxtimes v+1\boxtimes (v\otimes w)\end{aligned}}}$

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

The extension above preserves the algebra grading. That is,

${\displaystyle \Delta :T^{2}V\to \bigoplus _{k=0}^{2}T^{k}V\boxtimes T^{(2-k)}V}$

Continuing in this fashion, one can obtain an explicit expression for the coproduct acting on a homogenous element of order m:

${\displaystyle {\begin{aligned}\Delta (v_{1}\otimes \cdots \otimes v_{m})&=\Delta (v_{1})\otimes \cdots \otimes \Delta (v_{m})\\&=\sum _{p=0}^{m}\left(v_{0}\otimes \cdots \otimes v_{p}\right)\;\omega \;\left(v_{p+1}\otimes \cdots \otimes v_{m}\right)\\&=\sum _{p=0}^{m}\;\sum _{\sigma \in \mathrm {Sh} (p,m-p+1)}\;\left(v_{\sigma (0)}\otimes \dots \otimes v_{\sigma (p)}\right)\boxtimes \left(v_{\sigma (p+1)}\otimes \dots \otimes v_{\sigma (m)}\right)\end{aligned}}}$

where the ${\displaystyle \omega }$ 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,m-p+1)-shuffles. The above is written with a notational trick, to keep track of the field element 1: the trick is to write ${\displaystyle v_{0}=1}$, and this is shuffled into various locations during the expansion of the sum over shuffles. The shuffle follows directly from the first axiom of a co-algebra: the relative order of the elements ${\displaystyle v_{k}}$ 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. Any one given shuffle obeys

${\displaystyle \sigma (0)<\cdots <\sigma (p)\quad {\mbox{ and }}\quad \sigma (p+1)<\cdots <\sigma (m)}$

As before, the algebra grading is preserved:

${\displaystyle \Delta :T^{m}V\to \bigoplus _{k=0}^{m}T^{k}V\boxtimes T^{(m-k)}V}$

The counit ${\displaystyle \epsilon :TV\to K}$ is given by the projection of the field component out from the algebra. This can be written as ${\displaystyle \epsilon :v\mapsto 0}$ for ${\displaystyle v\in V}$ and ${\displaystyle \epsilon :k\mapsto k}$ for ${\displaystyle k\in K=T^{0}V}$. By homomorphism under the tensor product ${\displaystyle \otimes }$, this extends to

${\displaystyle \epsilon :x\mapsto 0}$

for all ${\displaystyle x\in T^{1}V\oplus T^{2}V\oplus \cdots }$ It is a straight-forward matter to verify that this counit satisfies the needed axiom for the coalgebra:

${\displaystyle (\mathrm {id} \boxtimes \epsilon )\circ \Delta =\mathrm {id} =(\epsilon \boxtimes \mathrm {id} )\circ \Delta .}$

Working this explicitly, one has

${\displaystyle {\begin{aligned}((\mathrm {id} \boxtimes \epsilon )\circ \Delta )(x)&=(\mathrm {id} \boxtimes \epsilon )(1\boxtimes x+x\boxtimes 1)\\&=1\boxtimes \epsilon (x)+x\boxtimes \epsilon (1)\\&=0+x\boxtimes 1\\&\cong x\end{aligned}}}$

where, for the last step, one has made use of the isomorphism ${\displaystyle TV\boxtimes K\cong TV}$, as is appropriate for the defining axiom of the counit.

Multiplication is given by an operator

${\displaystyle \nabla :TV\boxtimes TV\to TV}$

which, in this case, was already given as the "internal" tensor product. That is,

${\displaystyle \nabla :x\boxtimes y\mapsto x\otimes y}$

That is, ${\displaystyle \nabla (x\boxtimes y)=x\otimes y.}$ The above should make it clear why the ${\displaystyle \boxtimes }$ symbol needs to be used: the ${\displaystyle \otimes }$ was actually one and the same thing as ${\displaystyle \nabla }$; and notational sloppiness here would lead to utter chaos. To strengthen this: the tensor product ${\displaystyle \otimes }$ of the tensor algebra corresponds to the multiplication ${\displaystyle \nabla }$ used in the definition of an algebra, whereas the tensor product ${\displaystyle \boxtimes }$ is the one required in the definition of comultiplication in a coalgebra. These two tensor products are not the same thing!

The unit for the algebra

${\displaystyle \eta :K\to TV}$

is just the embedding, so that

${\displaystyle \eta :k\mapsto k}$

That the unit is compatible with the tensor product ${\displaystyle \otimes }$ is "trivial": it is just part of the standard definition of the tensor product of vector spaces. That is, ${\displaystyle k\otimes x=kx}$ for field element k and any ${\displaystyle x\in TV.}$ More verbosely, the axioms for an associative algebra require the two homomorphisms (or commuting diagrams):

${\displaystyle \nabla \circ (\eta \boxtimes \mathrm {id} _{TV})=\eta \otimes \mathrm {id} _{TV}=\eta \cdot \mathrm {id} _{TV}}$

on ${\displaystyle K\boxtimes TV}$, and that symmetrically, on ${\displaystyle TV\boxtimes K}$, that

${\displaystyle \nabla \circ (\mathrm {id} _{TV}\boxtimes \eta )=\mathrm {id} _{TV}\otimes \eta =\mathrm {id} _{TV}\cdot \eta }$

where the right-hand side of these equations should be understood as the scalar product.

The unit and counit, and multiplication and comultiplication, all have to satisfy compatibility conditions. It is straightforward to see that

${\displaystyle \epsilon \circ \eta =\mathrm {id} _{K}.}$

Similarly, the unit is compatible with comultiplication:

${\displaystyle \Delta \circ \eta =\eta \boxtimes \eta \cong \eta }$

The above requires the use of the isomorphism ${\displaystyle K\boxtimes K\cong K}$ in order to work; without this, one loses linearity. Component-wise,

${\displaystyle (\Delta \circ \eta )(k)=\Delta (k)=k(1\boxtimes 1)\cong k}$

with the right-hand side making use of the isomorphism.

Multiplication and the counit are compatible:

${\displaystyle (\epsilon \circ \nabla )(x\boxtimes y)=\epsilon (x\otimes y)=0}$

whenever x or y are not elements of ${\displaystyle K}$, and otherwise, one has scalar multiplication on the field: ${\displaystyle k_{1}\otimes k_{2}=k_{1}k_{2}.}$ The most difficult to verify is the compatibility of multiplication and comultiplication:

${\displaystyle \Delta \circ \nabla =(\nabla \boxtimes \nabla )\circ (\mathrm {id} \boxtimes \tau \boxtimes \mathrm {id} )\circ (\Delta \boxtimes \Delta )}$

where ${\displaystyle \tau (x\boxtimes y)=y\boxtimes x}$ exchanges elements. The compatibility condition only needs to be verified on ${\displaystyle V\subset TV}$; the full compatibility follows as a homomorphic extension to all of ${\displaystyle TV.}$ The verification is verbose but straight-forward; it is not given here, except for the final result:

${\displaystyle (\Delta \circ \nabla )(v\boxtimes w)=\Delta (v\otimes w)}$

For ${\displaystyle v,w\in V,}$ an explicit expression for this was given in the coalgebra section, above.

Compatibility of the antipode with multiplication and comultiplication requires that

${\displaystyle \nabla \circ (S\boxtimes \mathrm {id} )\circ \Delta =\eta \circ \epsilon =\nabla \circ (\mathrm {id} \boxtimes S)\circ \Delta }$

This is straight-forward to verify componentwise on ${\displaystyle k\in K}$:

${\displaystyle {\begin{aligned}(\nabla \circ (S\boxtimes \mathrm {id} )\circ \Delta )(k)&=(\nabla \circ (S\boxtimes \mathrm {id} ))(1\boxtimes k)\\&=\nabla (1\boxtimes k)\\&=1\otimes k\\&=k\end{aligned}}}$

Similarly, on ${\displaystyle v\in V}$:

${\displaystyle {\begin{aligned}(\nabla \circ (S\boxtimes \mathrm {id} )\circ \Delta )(v)&=(\nabla \circ (S\boxtimes \mathrm {id} ))(v\boxtimes 1+1\boxtimes v)\\&=\nabla (-v\boxtimes 1+1\boxtimes v)\\&=-v\otimes 1+1\otimes v\\&=-v+v\\&=0\end{aligned}}}$

Recall that

${\displaystyle (\eta \circ \epsilon )(k)=\eta (k)=k}$

and that

${\displaystyle (\eta \circ \epsilon )(x)=\eta (0)=0}$

for any ${\displaystyle x\in TV}$ that is not in ${\displaystyle K.}$

One may proceed in a similar manner, by homomorphism, verifying that the antipode inserts the appropriate cancellative signs in the shuffle, starting with the compatibility condition on ${\displaystyle T^{2}V}$ and proceeding by induction.