Opposite Algebras via automorphism of PROP

Mirror Symmetry and Opposite Algebras

Let $A$$A$ be an associative algebra. Its opposite algebra $A^{\mathrm{op}}$$A^{\mathrm{op}}$ has the same underlying object, with multiplication

$${\mu }^{\mathrm{op}}=\mu \circ \tau .$$

In element notation,

$$x{\cdot }_{\mathrm{op}}y=y\cdot x.$$

It is easy to verify directly that this multiplication is again associative and has the same unit. But the formula suggests a more structural question:

Why does reversing the inputs of multiplication produce another algebra of exactly the same kind?

The answer becomes clearer at the level of PROPs.

There are two different ways to identify a PROP with its horizontal mirror. One is geometric and exists for every PROP. The other depends on the chosen algebraic operations and their relations. The opposite operation appears by comparing these two identifications.

The mirror PROP

Let $P$$P$ be a PROP. Its objects are the natural numbers, with

$$m\otimes n=m+n.$$

It is useful to regard

$$n=1\otimes \cdots \otimes 1$$

as a row of $n$$n$ ordered slots,

$$x_1\otimes \cdots \otimes x_n.$$

Define $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$ to have the same underlying category as $P$$P$, but with the tensor product reversed:

$$f{\otimes }_{\mathrm{mir}}g:=g\otimes f.$$

On objects,

$$m{\otimes }_{\mathrm{mir}}n:=n+m.$$

Since the objects of a PROP are natural numbers,

$$n+m=m+n,$$

so this is again a strict monoidal structure.

The unit object is still $0$$0$, and the symmetry in $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$ is defined by

$${\tau }_{m,n}^{\mathrm{mir}}:={\tau }_{n,m}.$$

Thus $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$ is again a PROP.

This is the reverse monoidal structure on $P$$P$. The notation $P^{\mathrm{rev}}$$P^{\mathrm{rev}}$ is also standard; here we use $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$ to emphasize the picture of horizontal reflection.

The geometric mirror

For each $n$$n$, let

$${\rho }_n:n⟶n$$

be the permutation reversing the order of the $n$$n$ slots:

$$x_1\otimes \cdots \otimes x_n⟼x_n\otimes \cdots \otimes x_1.$$

Thus

$${\rho }_0=1_0,{\rho }_1=1_1,{\rho }_2={\tau }_{1,1},$$

and

$${\rho }_n^2=1_n.$$

If a row is divided into two blocks

$$[A\mid B],$$

then reversing the whole row gives

$$[B^{\mathrm{rev}}\mid A^{\mathrm{rev}}].$$

Equivalently,

$${\rho }_{m+n}=({\rho }_n\otimes {\rho }_m)\circ {\tau }_{m,n}.$$

Now define

$$M_P:P⟶P^{\mathrm{mir}}$$

to be the identity on objects, and for $f:m\to n$$f:m\to n$ set

$$M_P(f):={\rho }_n\circ f\circ {\rho }_m.$$

This is the literal horizontal reflection of the string diagram of $f$$f$.

Vertical composition is preserved. If

$$f:m\to n,g:n\to k,$$

then

$$\begin{array}{rl}{M}_P(g\circ f)& ={\rho }_k\circ g\circ f\circ {\rho }_m\\ & ={\rho }_k\circ g\circ {\rho }_n\circ {\rho }_n\circ f\circ {\rho }_m\\ & =M_P(g)\circ M_P(f).\end{array}$$

For the tensor product, let

$$f:m\to n,g:p\to q.$$

Using

$${\rho }_{n+q}=({\rho }_q\otimes {\rho }_n)\circ {\tau }_{n,q},$$

together with

$${\tau }_{n,q}\circ (f\otimes g)=(g\otimes f)\circ {\tau }_{m,p},$$

we obtain

$$\begin{array}{rl}{M}_P(f\otimes g)& ={\rho }_{n+q}\circ (f\otimes g)\circ {\rho }_{m+p}\\ & =({\rho }_q\otimes {\rho }_n)\circ {\tau }_{n,q}\circ (f\otimes g)\circ {\rho }_{m+p}\\ & =({\rho }_q\otimes {\rho }_n)\circ (g\otimes f)\circ {\tau }_{m,p}\circ {\rho }_{m+p}\\ & =({\rho }_q\otimes {\rho }_n)\circ (g\otimes f)\circ ({\rho }_p\otimes {\rho }_m)\\ & =M_P(g)\otimes M_P(f).\end{array}$$

Since

$$M_P(f){\otimes }_{\mathrm{mir}}{M}_P(g)=M_P(g)\otimes M_P(f),$$

this says

$$M_P(f\otimes g)=M_P(f){\otimes }_{\mathrm{mir}}{M}_P(g).$$

Hence $M_P$$M_P$ is strict monoidal.

It is also symmetric monoidal. Indeed,

$$M_P({\tau }_{m,n})={\rho }_{m+n}\circ {\tau }_{m,n}\circ {\rho }_{m+n}={\tau }_{n,m}={\tau }_{m,n}^{\mathrm{mir}}.$$

Therefore

$$\boxed{{\displaystyle M_P:P\stackrel{\sim }{⟶}{P}^{\mathrm{mir}}}}$$

is a strict symmetric monoidal isomorphism.

The reversal permutation defined in $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$ is again the same underlying morphism ${\rho }_n$$\rho_n$, so reflecting twice gives

$$M_{P^{\mathrm{mir}}}\circ M_P={\mathrm{id}}_P.$$

Thus every PROP is canonically isomorphic to its horizontal mirror.

This fact is universal. It is not yet the special symmetry behind opposite algebras.

Local reflection in the free PROP

Now choose a signature $\mathrm{\Sigma }$$\Sigma$ of basic operations and let

$$F_{\mathrm{\Sigma }}:=\mathsf{FPROP}(\mathrm{\Sigma })$$

be the free PROP generated by $\mathrm{\Sigma }$$\Sigma$.

For each generator

$$x:m\to n,$$

define its local reflection by

$$x⟼{\rho }_n\circ x\circ {\rho }_m.$$

By the universal property of the free PROP, this assignment extends uniquely to a strict symmetric monoidal endofunctor

$${\mathrm{\Omega }}_{\mathrm{\Sigma }}:F_{\mathrm{\Sigma }}⟶F_{\mathrm{\Sigma }}$$

such that

$$\boxed{{\displaystyle {\mathrm{\Omega }}_{\mathrm{\Sigma }}(x)={\rho }_n\circ x\circ {\rho }_m.}}$$

Because ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$$\Omega_\Sigma$ is symmetric monoidal and is the identity on objects, it fixes all permutation morphisms. In particular,

$${\mathrm{\Omega }}_{\mathrm{\Sigma }}({\rho }_n)={\rho }_n.$$

Hence, for every generator $x:m\to n$$x:m\to n$,

$$\begin{array}{rl}{\mathrm{\Omega }}_{\mathrm{\Sigma }}^2(x)& ={\mathrm{\Omega }}_{\mathrm{\Sigma }}({\rho }_n\circ x\circ {\rho }_m)\\ & ={\rho }_n\circ ({\rho }_n\circ x\circ {\rho }_m)\circ {\rho }_m\\ & =x.\end{array}$$

By uniqueness of the extension from the free PROP,

$$\boxed{{\displaystyle {\mathrm{\Omega }}_{\mathrm{\Sigma }}^2={\mathrm{id}}_{F_{\mathrm{\Sigma }}}.}}$$

Thus local reflection is already an involution on the free theory.

Relations and mirror symmetry

Now let

$$P=F_{\mathrm{\Sigma }}/{\sim }_R,$$

where $R$$R$ is a collection of equations between parallel morphisms and ${\sim }_R$$\sim_R$ is the PROP congruence generated by those equations.

Let

$$q:F_{\mathrm{\Sigma }}⟶P$$

be the quotient map.

The local reflection ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$$\Omega_\Sigma$ descends to $P$$P$ precisely when

$$u{\sim }_{R}v⟹{\mathrm{\Omega }}_{\mathrm{\Sigma }}(u){\sim }_R{\mathrm{\Omega }}_{\mathrm{\Sigma }}(v).$$

Equivalently,

$${\mathrm{\Omega }}_{\mathrm{\Sigma }}({\sim }_R)\subseteq {\sim }_R.$$

Since ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$$\Omega_\Sigma$ is an involution, this inclusion is equivalent to equality:

$$\boxed{{\displaystyle {\mathrm{\Omega }}_{\mathrm{\Sigma }}({\sim }_R)={\sim }_R.}}$$

If $⟨R⟩$$\langle R\rangle$ denotes the congruence generated by $R$$R$, the same condition can be written as

$$\boxed{{\displaystyle {\mathrm{\Omega }}_{\mathrm{\Sigma }}⟨R⟩=⟨R⟩.}}$$

This is the precise mirror-symmetry condition.

In practice one does not need to check every equation in the generated congruence. It is enough to check that, for every defining equation

$$u=v$$

in $R$$R$, the reflected equation

$${\mathrm{\Omega }}_{\mathrm{\Sigma }}(u)={\mathrm{\Omega }}_{\mathrm{\Sigma }}(v)$$

is derivable from $R$$R$.

When the congruence is invariant, ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$$\Omega_\Sigma$ descends to a strict symmetric monoidal endofunctor

$$\mathrm{\Omega }:P⟶P.$$

Since ${\mathrm{\Omega }}_{\mathrm{\Sigma }}^2=1$$\Omega_\Sigma^2=1$, the descended functor satisfies

$$\boxed{{\displaystyle {\mathrm{\Omega }}^2={\mathrm{id}}_P.}}$$

Hence $\mathrm{\Omega }$$\Omega$ is a strict symmetric monoidal automorphism.

A second identification with the mirror

There is another way to see the same condition.

Regard every chosen generator

$$x:m\to n$$

as the same underlying morphism in $P^{\mathrm{mir}}$$P^{\mathrm{mir}}$.

By the universal property of the free PROP, there is a unique strict symmetric monoidal functor

$$\tilde{J}:F_{\mathrm{\Sigma }}⟶P^{\mathrm{mir}}$$

such that

$$\tilde{J}(x)=q(x)$$

for every $x\in \mathrm{\Sigma }$$x\in\Sigma$.

Because $\tilde{J}$$\widetilde J$ is strict monoidal,

$$\tilde{J}(f\otimes g)=\tilde{J}(f){\otimes }_{\mathrm{mir}}\tilde{J}(g),$$

and because it is symmetric monoidal,

$$\tilde{J}({\tau }_{m,n})={\tau }_{m,n}^{\mathrm{mir}}.$$

Thus $\tilde{J}$$\widetilde J$ keeps the basic operations fixed, but reverses their horizontal arrangement.

The key identity is

$$\boxed{{\displaystyle \tilde{J}=M_P\circ q\circ {\mathrm{\Omega }}_{\mathrm{\Sigma }}.}}$$

Indeed, for a generator $x:m\to n$$x:m\to n$,

$$\begin{array}{rl}{M}_{P}q{\mathrm{\Omega }}_{\mathrm{\Sigma }}(x)& =M_P({\rho }_n\circ q(x)\circ {\rho }_m)\\ & ={\rho }_n\circ ({\rho }_n\circ q(x)\circ {\rho }_m)\circ {\rho }_m\\ & =q(x).\end{array}$$

Both sides are strict symmetric monoidal functors out of $F_{\mathrm{\Sigma }}$$F_\Sigma$ and agree on every generator, so they are equal.

It follows that $\tilde{J}$$\widetilde J$ descends to

$$J:P⟶P^{\mathrm{mir}}$$

if and only if ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$$\Omega_\Sigma$ preserves the defining congruence.

Therefore

$$\boxed{{\displaystyle J\text{ exists}⟺{\mathrm{\Omega }}_{\mathrm{\Sigma }}({\sim }_R)={\sim }_R.}}$$

When this condition holds, the identity above descends to

$$\boxed{{\displaystyle J=M_P\circ \mathrm{\Omega }.}}$$

Since both $M_P$$M_P$ and $\mathrm{\Omega }$$\Omega$ are strict symmetric monoidal isomorphisms, so is $J$$J$.

We therefore obtain

$$\mathrm{\Omega }=M_P^{-1}\circ J.$$

Since ${\mathrm{\Omega }}^{-1}=\mathrm{\Omega }$$\Omega^{-1}=\Omega$, this is equivalently

$$\boxed{{\displaystyle \mathrm{\Omega }=J^{-1}\circ M_P.}}$$

Thus the internal opposite involution is the discrepancy between two identifications of $P$$P$ with its mirror.

Associative algebras

Consider the PROP $P_{\mathrm{Ass}}$$P_{\mathrm{Ass}}$ governing associative unital algebras. It is generated by

$$\mu :2\to 1,\eta :0\to 1,$$

subject to associativity

$$\mu \circ (\mu \otimes 1)=\mu \circ (1\otimes \mu )$$

and the unit equations

$$\mu \circ (\eta \otimes 1)=1,$$

$$\mu \circ (1\otimes \eta )=1.$$

The mirror symmetry is particularly visible here.

Keep the generators $\mu$$\mu$ and $\eta$$\eta$ fixed and interpret the same equations using the mirror tensor.

Associativity becomes

$$\mu \circ (\mu {\otimes }_{\mathrm{mir}}1)=\mu \circ (1{\otimes }_{\mathrm{mir}}\mu ).$$

But

$$\mu {\otimes }_{\mathrm{mir}}1=1\otimes \mu ,$$

while

$$1{\otimes }_{\mathrm{mir}}\mu =\mu \otimes 1.$$

Hence the mirrored associativity equation is

$$\mu \circ (1\otimes \mu )=\mu \circ (\mu \otimes 1),$$

which is exactly the original equation with the two sides exchanged.

One can already see the symmetry in the identity

$$\mu \circ (\mu \otimes 1)=\mu \circ (\mu {\otimes }_{\mathrm{mir}}1).$$

The unit equations behave in the same way:

$$\eta {\otimes }_{\mathrm{mir}}1=1\otimes \eta ,$$

and

$$1{\otimes }_{\mathrm{mir}}\eta =\eta \otimes 1.$$

Thus the left and right unit equations are exchanged.

Therefore the congruence defining $P_{\mathrm{Ass}}$$P_{\mathrm{Ass}}$ is invariant under local reflection. Hence there is a strict symmetric monoidal isomorphism

$$\boxed{{\displaystyle J:P_{\mathrm{Ass}}\stackrel{\sim }{⟶}{P}_{\mathrm{Ass}}^{\mathrm{mir}}}}$$

which fixes the chosen generators:

$$J(\mu )=\mu ,J(\eta )=\eta .$$

At the same time, the internal involution satisfies

$$\mathrm{\Omega }(\mu )={\rho }_1\circ \mu \circ {\rho }_2=\mu \circ \tau ,$$

and

$$\mathrm{\Omega }(\eta )=\eta .$$

Thus

$$\boxed{{\displaystyle \mathrm{\Omega }(\mu )=\mu \circ \tau .}}$$

This is exactly the opposite multiplication.

The opposite algebra

Let

$$A:P_{\mathrm{Ass}}⟶\mathcal{C}$$

be an associative algebra in a symmetric monoidal category $\mathcal{C}$$\mathcal C$.

Precomposing with $\mathrm{\Omega }$$\Omega$ gives another algebra

$$A^{\mathrm{op}}:=A\circ \mathrm{\Omega }.$$

Its multiplication is

$$A(\mathrm{\Omega }(\mu ))=A(\mu \circ \tau )={\mu }_A\circ {\tau }_{A,A}.$$

Hence

$$x{\cdot }_{\mathrm{op}}y=y\cdot x.$$

The usual opposite algebra therefore comes from comparing two different identifications

$$P_{\mathrm{Ass}}\simeq P_{\mathrm{Ass}}^{\mathrm{mir}}.$$

The first is the universal geometric mirror

$$M_P.$$

The second is the symmetry of the chosen associative presentation

$$J.$$

Their discrepancy is

$$\boxed{{\displaystyle \mathrm{\Omega }=J^{-1}\circ M_P,}}$$

and this sends

$$\mu ⟼\mu \circ \tau .$$

The opposite multiplication is therefore not an isolated trick. It is the internal trace of a mirror symmetry of the presentation.

Coalgebras

The same argument applies to the PROP governing coassociative counital coalgebras.

The basic operations are

$$\mathrm{\Delta }:1\to 2,ϵ:1\to 0.$$

The geometric mirror sends

$$\mathrm{\Delta }⟼{\rho }_2\circ \mathrm{\Delta }\circ {\rho }_1=\tau \circ \mathrm{\Delta }.$$

Coassociativity

$$(\mathrm{\Delta }\otimes 1)\circ \mathrm{\Delta }=(1\otimes \mathrm{\Delta })\circ \mathrm{\Delta }$$

is invariant under replacing $\otimes$$\otimes$ by ${\otimes }_{\mathrm{mir}}$$\otimes_{\mathrm{mir}}$, while the left and right counit equations are exchanged.

Hence the coalgebra presentation also has mirror symmetry, and the associated internal involution sends

$$\boxed{{\displaystyle \mathrm{\Delta }⟼\tau \circ \mathrm{\Delta }.}}$$

This is the usual coopposite coalgebra.

Two useful extremes

For commutative algebras,

$$\mu \circ \tau =\mu .$$

Therefore

$$\mathrm{\Omega }=\mathrm{id}.$$

Consequently,

$$J=M_P.$$

In the commutative theory, geometric reflection and the generator-fixing mirror identification coincide.

At the opposite extreme, consider a theory generated by

$$\mu :2\to 1,\eta :0\to 1,$$

with only the left unit equation

$$\mu \circ (\eta \otimes 1)=1.$$

Under horizontal reversal this becomes

$$\mu \circ (1\otimes \eta )=1,$$

the right unit equation.

The right unit equation is not a consequence of the left unit equation in general. Hence the defining congruence is not invariant under local reflection.

Therefore $M_P$$M_P$ still exists, as it does for every PROP, but no generator-fixing

$$J:P⟶P^{\mathrm{mir}}$$

exists.

This sharply separates the two notions:

$$\boxed{{\displaystyle M_P\text{ is universal, while }J\text{ records a symmetry of the chosen presentation.}}}$$

Reverse monoidal categories

There is also a semantic version of the same picture.

For a monoidal category $\mathcal{C}$$\mathcal C$, define the reverse monoidal category ${\mathcal{C}}^{\mathrm{rev}}$$\mathcal C^{\mathrm{rev}}$ by

$$X{\otimes }_{\mathrm{rev}}Y:=Y\otimes X.$$

Then one has the familiar correspondence

$${\mathrm{Fun}}^{\otimes }(P^{\mathrm{mir}},\mathcal{C})\cong {\mathrm{Fun}}^{\otimes }(P,{\mathcal{C}}^{\mathrm{rev}}).$$

If $\mathcal{C}$$\mathcal C$ is symmetric monoidal, its symmetry gives a strong symmetric monoidal equivalence

$$(\mathrm{id},\tau ):\mathcal{C}\simeq {\mathcal{C}}^{\mathrm{rev}}.$$

The geometric mirror $M_P$$M_P$ is the syntactic PROP-level counterpart of this equivalence. On tensor powers, the iterated symmetry is exactly the reversal permutation

$${\rho }_n:A^{\otimes n}⟶A^{\otimes n}.$$

When a presentation admits $J$$J$, the same chosen operations satisfy the same theory after the ambient tensor order is reversed. Transporting back along the symmetry of $\mathcal{C}$$\mathcal C$ gives the opposite structure.

The PROP involution $\mathrm{\Omega }$$\Omega$ is the syntactic form of this construction.

What the symmetry depends on

There is one final qualification.

Mirror symmetry in this sense is not a property of the bare PROP $P$$P$ alone. It depends on a chosen collection of basic operations

$$\mathrm{\Sigma }⟶P.$$

Once these operations are fixed, the property depends only on the congruence they satisfy, not on a particular list of equations generating that congruence.

Thus replacing $R$$R$ by another set of equations generating the same congruence does not change the answer.

Changing the chosen generators can change the answer.

The relevant object is therefore better thought of as a PROP equipped with chosen basic operations, or equivalently a marked presentation.

For

$$P=F_{\mathrm{\Sigma }}/{\sim }_R,$$

the precise criterion is

$$\boxed{{\displaystyle {\mathrm{\Omega }}_{\mathrm{\Sigma }}({\sim }_R)={\sim }_R.}}$$

Equivalently, the mirror of every defining equation must be derivable from the original theory.

When this happens, the chosen presentation has a genuine mirror symmetry

$$J:P\stackrel{\sim }{⟶}{P}^{\mathrm{mir}},$$

and comparing it with the universal geometric mirror produces the internal opposite involution

$$\mathrm{\Omega }=J^{-1}\circ M_P.$$