$\mathsf{S}(\mathcal{P},-):\mathsf{MOD}_{R}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{MOD}_{R}$ which maps an
$R$-module $V$ to $\mathsf{S}(\mathcal{P},V)$, and sends an $R$-linear
morphism ${f:V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}W}$ to the $R$-linear morphism
${\mathsf{S}(\mathcal{P},f):\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},W)}$ which is
defined as follows:

$\displaystyle\mathsf{S}(\mathcal{P},f)(\mu;v_{1},\dots,v_{n})=(\mu;f(v_{1}),\dots,f(v_{n}))$

The monad unit $\eta_{V}:V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V)$ and the monad
multiplication
$\gamma_{V}:S\left(\mathcal{P},\mathsf{S}(\mathcal{P},V)\right)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V)$ are
respectively defined as follows:

$\eta_{V}(v)=(1_{\mathcal{P}};v)\in\mathcal{P}(1)\otimes V,$
$\gamma_{V}\left(\mu;\left(\nu_{1};v_{1,1},\ldots,v_{1,n_{1}}\right),\ldots,\left(\nu_{k};v_{k,1},\ldots,v_{k,n_{k}}\right)\right)=\left(\mu\left(\nu_{1},\ldots,\nu_{k}\right);v_{1,1},\ldots,v_{k,n_{k}}\right).$

Then, $(\mathsf{S}(\mathcal{P},-),\gamma,\eta)$ is a monad on
$\mathsf{MOD}_{R}$ [26, Section 3]. We will now prove that this monad is in
fact also a coCartesian differential monad.

######  Theorem 4.1.1.

Let $\mathcal{P}$ be an operad. Let
$\partial_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V\times V)$ be
the $R$-linear map such that:

$\begin{gathered}\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)=\sum_{i=1}^{n}\left(\mu;(v_{1},0),\ldots,(0,v_{i}),\ldots,(v_{n},0)\right).\end{gathered}$

Here, in the sum, we use the first injection
$V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}V\times V$ to all the inputs in $V$
except for the $i$-th input, for which we use the second injection. Then,
$\partial$ is a differential combinator transformation for
$(\mathsf{S}(\mathcal{P},-),\gamma,\eta)$, and thus
$(\mathsf{S}(\mathcal{P},-),\gamma,\eta,\partial)$ is a coCartesian
differential monad.

###### Proof.

It is clear that $\partial$ is a natural transformation. Therefore, we must
prove that [DC.1] to [DC.6] in Definition 3.2.1 hold. First observe that for
any $R$-linear morphism $f:V\times V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}W$, we have:

$\displaystyle\mathsf{S}(\mathcal{P},f)\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)=\sum_{i=1}^{n}\left(\mu;f(v_{1},0),\ldots,f(0,v_{i}),\ldots,f(v_{n},0)\right).$

This will help simplify our calculations.

  1. [DC.1]

Here, we use the fact that $(\mu;v_{1},\ldots,0,\ldots,v_{n})=0$:

$\displaystyle\mathsf{S}(\mathcal{P},\pi_{1})\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)=\sum_{i=1}^{n}\left(\mu;v_{1},\ldots,0,\ldots,v_{n}\right)=0$

So $\mathsf{S}(\mathcal{P},\pi_{1})\circ\partial_{V}=0$

  2. [DC.2]

Here we use that
$(\mu;v_{1},\ldots,v_{i}+v_{i}^{\prime},\ldots,v_{n})=(\mu;v_{1},\ldots,v_{i},\ldots,v_{n})+(\mu;v_{1},\ldots,v_{i}^{\prime},\ldots,v_{n})$:

$\mathsf{S}(\mathcal{P},\langle\pi_{1},\pi_{2},\pi_{2}\rangle)\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)=\sum_{i=1}^{n}\left(\mu;(v_{1},0,0),\ldots,(0,v_{i},v_{i}),\ldots,(v_{n},0,0)\right)\\\
=\sum_{i=1}^{n}\left(\mu;(v_{1},0,0),\ldots,(0,v_{i},0),\ldots,(v_{n},0,0)\right)\\\
\qquad+\sum_{i=1}^{n}\left(\mu;(v_{1},0,0),\ldots,(0,0,v_{i}),\ldots,(v_{n},0,0)\right)\\\
=\mathsf{S}(\mathcal{P},\langle\pi_{1},\pi_{2},0\rangle)\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)+\mathsf{S}(\mathcal{P},\langle\pi_{1},0,\pi_{2}\rangle)\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right),$

so
$\mathsf{S}(\mathcal{P},\langle\pi_{1},\pi_{2},\pi_{2}\rangle)\circ\partial_{V}=\mathsf{S}(\mathcal{P},\langle\pi_{1},\pi_{2},0\rangle)\circ\partial_{V}+\mathsf{S}(\mathcal{P},\langle\pi_{1},0,\pi_{2}\rangle)\circ\partial_{V}$

  3. [DC.3]

This is just the case for $n=1$ which says that
$\partial_{V}(\mu;v)=(\mu;(0,v))$. So $\partial_{V}\circ\eta_{V}=\eta_{V\times
V}\circ\langle 0,1_{V}\rangle$.

  4. [DC.4]

We can start with an element of the form
$\left(\mu;\left(\nu_{1};v_{1,1},\ldots,v_{1,n_{1}}\right),\ldots,\left(\nu_{k};v_{k,1},\ldots,v_{k,n_{k}}\right)\right)$.
On the one hand, we have that:

$\partial_{V}\left(\gamma_{V}\left(\mu;\left(\nu_{1};v_{1,1},\ldots,v_{1,n_{1}}\right),\ldots,\left(\nu_{k};v_{k,1},\ldots,v_{k,n_{k}}\right)\right)\right)\\\
=\partial_{V}\left(\left(\mu\left(\nu_{1},\ldots,\nu_{k}\right);v_{1,1},\ldots,v_{k,n_{k}}\right)\right)\\\
=\sum\limits^{k}_{i=1}\sum^{n_{i}}_{j_{i}=1}\left(\mu\left(\nu_{1},\ldots,\nu_{k}\right);(v_{1,1},0),\ldots,(0,v_{i,j_{i}}),\ldots,(v_{k,n_{k}},0)\right).$

On the other hand, we have that:

$\partial_{\mathsf{S}(\mathcal{P},V)}\left(\mu;\left(\nu_{1};v_{1,1},\ldots,v_{1,n_{1}}\right),\ldots,\left(\nu_{k};v_{k,1},\ldots,v_{k,n_{k}}\right)\right)\\\
\hskip-11.38092pt=\sum\limits^{k}_{i=1}\Bigg{(}\mu;\left(\left(\nu_{1};v_{1,1},\ldots,v_{1,n_{1}}\right),0\right),\ldots,\left(0,\left(\nu_{i};v_{i,1},\ldots,v_{i,n_{i}}\right)\right),\ldots\left(\left(\nu_{k};v_{k,1},\ldots,v_{k,n_{k}}\right),0\right)\Bigg{)}.$

Then applying $\mathsf{S}\left(\mathcal{P},\mathsf{S}(\mathcal{P},\langle
1_{V},0\rangle)\circ\pi_{1}+\partial_{V}\circ\pi_{2}\right)$, using the
multilinearity in the module arguments, we get:

$\sum\limits^{k}_{i=1}\sum^{n_{i}}_{j_{i}=1}\bigg{(}\mu;\left(\nu_{1};(v_{1,1},0),\ldots,(v_{1,n_{1}},0)\right),\\\
\ldots,\left(\nu_{i};(v_{i,1},0),\ldots,(0,v_{i,j_{i}}),\ldots,(v_{i,n_{i}},0)\right),\ldots,\left(\nu_{k};(v_{k,1},0),\ldots,(v_{k,n_{k}},0)\right)\bigg{)}.$

Then, finally applying $\gamma_{V\times V}$, we get:

$\displaystyle\sum\limits^{k}_{i=1}\sum^{n_{i}}_{j_{i}=1}\left(\mu\left(\nu_{1},\ldots,\nu_{k}\right);(v_{1,1},0),\ldots,(0,v_{i,j_{i}}),\ldots,(v_{k,n_{k}},0)\right).$

So, $\partial_{V}\circ\gamma_{V}=\gamma_{V\times V}\circ
S\left(\mathcal{P},\mathsf{S}(\mathcal{P},\langle
1_{V},0\rangle)\circ\pi_{1}+\partial_{V}\circ\pi_{2}\right)\circ\partial_{\mathsf{S}(\mathcal{P},V)}$.

For the remaining two relations, let us first expand out $\partial_{V\times
V}\circ\partial_{V}$:

$\partial_{V\times
V}\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)\\\
=\sum_{i=1}^{n}\sum_{1\leq
j<i}\left(\mu;(v_{1},0,0,0),\ldots,(0,0,v_{j},0)\ldots,(0,v_{i},0,0),\ldots,(v_{n},0,0,0)\right)\\\
+\sum_{i=1}^{n}\left(\mu;(v_{1},0,0,0),\ldots,(0,0,0,v_{i})\ldots,,(v_{n},0,0,0)\right)\\\
+\sum_{i=1}^{n}\sum_{i<j\leq
n}\left(\mu;(v_{1},0,0,0),\ldots,(0,v_{i},0,0),\ldots,(0,0,v_{j},0),\ldots,(v_{n},0,0,0)\right).$

  1. [DC.5]

Here we again use that $(\mu;v_{1},\ldots,0,\ldots,v_{n})=0$:

$\mathsf{S}(\langle\pi_{1},\pi_{4}\rangle)\left(\partial_{V\times
V}\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)\right)\\\
=\sum_{i=1}^{n}\sum_{1\leq
j<i}\left(\mu;(v_{1},0),\ldots,(0,0)\ldots,(0,0),\ldots,(v_{n},0)\right)\\\
+\sum_{i=1}^{n}\left(\mu;(v_{1},0),\ldots,(0,v_{i})\ldots,,(v_{n},0)\right)\\\
+\sum_{i=1}^{n}\sum_{i<j\leq
n}\left(\mu;(v_{1},0),\ldots,(0,0),\ldots,(0,0),\ldots,(v_{n},0,0,0)\right)\\\
=0+\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)+0=\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right),$

so $\mathsf{S}(\langle\pi_{1},\pi_{4}\rangle)\circ\partial_{V\times
V}\circ\partial_{V}=\partial_{V}$.

  2. [DC.6]

This amounts to a simple reindexing by swapping $i$ and $j$:

$\displaystyle
S\left(\mathcal{P},\left\langle\pi_{1},\pi_{3},\pi_{2},\pi_{4}\right\rangle\right)\left(\partial_{V\times
V}\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right)\right)=$
$\displaystyle=\sum_{i=1}^{n}\sum_{1\leq
j<i}\left(\mu;(v_{1},0,0,0),\ldots,(0,v_{j},0,0)\ldots,(0,0,v_{i},0),\ldots,(v_{n},0,0,0)\right)$
$\displaystyle+\sum_{i=1}^{n}\left(\mu;(v_{1},0,0,0),\ldots,(0,0,0,v_{i})\ldots,,(v_{n},0,0,0)\right)$
$\displaystyle+\sum_{i=1}^{n}\sum_{i<j\leq
n}\left(\mu;(v_{1},0,0,0),\ldots,(0,v_{i},0,0),\ldots,(0,0,v_{j},0),\ldots,(v_{n},0,0,0)\right)$
$\displaystyle=\sum_{j=1}^{n}\sum_{1\leq
i<j}\left(\mu;(v_{1},0,0,0),\ldots,(0,0,v_{i},0)\ldots,(0,v_{j},0,0),\ldots,(v_{n},0,0,0)\right)$
$\displaystyle+\sum_{j=1}^{n}\left(\mu;(v_{1},0,0,0),\ldots,(0,0,0,v_{j})\ldots,,(v_{n},0,0,0)\right)$
$\displaystyle+\sum_{j=1}^{n}\sum_{j<i\leq
n}\left(\mu;(v_{1},0,0,0),\ldots,(0,v_{j},0,0),\ldots,(0,0,v_{i},0),\ldots,(v_{n},0,0,0)\right)$
$\displaystyle=\partial_{V\times
V}\left(\partial_{V}\left(\mu;v_{1},\ldots,v_{n}\right)\right),$

so
$S\left(\mathcal{P},\left\langle\pi_{1},\pi_{3},\pi_{2},\pi_{4}\right\rangle\right)\partial_{V\times
V}\circ\partial_{V}=\partial_{V\times V}\circ\partial_{V}$.

We conclude that $\partial$ is a differential combinator transformation and
that $(\mathsf{S}(\mathcal{P},-),\gamma,\eta,\partial)$ is a coCartesian
differential monad. ∎

Here are now some well-known examples of operads, their associated monad and
resulting differential combinator transformation:

######  Example 4.1.2.

Any (unital and associative) $R$-algebra $A$ induces an operad $A^{\bullet}$
where $A^{\bullet}(1)=A$ and $A^{\bullet}(n)=\mathsf{0}$ for $n\neq 1$. The
associated monad is given by the free (left) $A$-module monad444To avoid
confusion, the algebras of the operad $A^{\bullet}$ will simply be (left)
$A$-modules and not $A$-algebras, as we will see in Example 4.3.4., that is,
$\mathsf{S}(A^{\bullet},V)=A\otimes V$. See [28, Example 0, 5.2.10] for full
details. The differential combinator transformation $\partial_{V}:A\otimes
V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\otimes(V\times V)$ is simply given by
injecting $V$ into the second component, so $\partial_{V}(a\otimes
v)=a\otimes(0,v)$.

######  Example 4.1.3.

The operad ${\operatorname{Com}}$ is defined by ${\operatorname{Com}}(n)=R$
for all $n$, with trivial action of the symmetric group. Unit and compositions
are defined by identities of $R$. The associated monad is the symmetric
algebra monad:

$\mathsf{S}({\operatorname{Com}},V)=\mathsf{Sym}(V)=\bigoplus_{n\in\mathbb{N}}\left(V^{\otimes^{n}}\right)_{\Sigma(n)}$

See [28, Example 2, 5.2.10] for full details. The differential combinator
transformation $\partial_{V}:\mathsf{Sym}(V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Sym}(V\times V)$ is defined on
pure symmetrized tensors as follows:

$\partial_{V}([v_{1}\otimes\ldots\otimes
v_{n}])=\sum^{n}_{i=1}[(v_{1},0)\otimes\ldots\otimes(0,v_{i})\otimes\ldots\otimes(v_{n},0)].$

Now, if $V$ is a free $R$-module with basis $X$, then $\mathsf{Sym}(V)$ is
isomorphic as an $R$-algebra to the polynomial $R$-algebra over $X$,
$\mathsf{Sym}(V)\cong R[X]$. Also, $\mathsf{Sym}(V\times V)$ is isomorphic to
the polynomial $R$-algebra over the disjoint union of $X$ with itself. So,
writing $dX=\\{dx|~{}\forall x\in X\\}$ to distinguish between the first copy
and the second copy of $X$, we have that $\mathsf{Sym}(V\times V)\cong
R[X,dX]$. In terms of polynomials,
$\partial_{V}:R[X]\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}R[X,dX]$ maps a polynomial to the sum of
its partial derivatives:

$\partial_{V}(p(\vec{x}))=\sum^{n}_{i=1}\frac{\partial p(\vec{x})}{\partial
x_{i}}dx_{i}.$

Therefore, $\partial$ recaptures polynomial differentiation.

######  Example 4.1.4.

There is an operad ${\operatorname{Ass}}$ defined by
${\operatorname{Ass}}(n)=R[\Sigma(n)]$, the regular representation of the
group $\Sigma(n)$. See [28, Example 1, 5.2.10] for full details. The
associated monad is the tensor algebra monad:

$\mathsf{S}({\operatorname{Ass}},V)=\mathsf{Ten}(V)=\bigoplus_{n\in\mathbb{N}}V^{\otimes
n}.$

The differential combinator transformation
$\partial_{V}:\mathsf{Ten}(V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Ten}(V\times V)$ is defined on
pure tensors as follows:

$\partial_{V}(v_{1}\otimes\ldots\otimes
v_{n})=\sum^{n}_{i=1}(v_{1},0)\otimes\ldots\otimes(0,v_{i})\otimes\ldots\otimes(v_{n},0).$

Now, if $V$ is a free $R$-module with basis $X$, then $\mathsf{Ten}(V)$ is
isomorphic to the $R$-algebra of non-commutative polynomials over $X$. As such
$\partial$ corresponds to differentiating non-commutative polynomials.

######  Example 4.1.5.

There is an operad ${\operatorname{Lie}}$ whose associated monad is given by
the free Lie algebra monad,
$\mathsf{S}({\operatorname{Lie}},V)=\mathsf{Lie}(V)$. See [28, Example 3,
5.2.10] for full details. In particular, $\mathsf{Lie}(V)$ is spanned by
elements of the form:
$\left[v_{1},[v_{2},\ldots[v_{n-1},v_{n}]\ldots]\right]$, so the Lie bracket
of the Lie bracket of etc. of elements $v_{i}\in V$. The differential
combinator transformation
$\partial_{V}:\mathsf{Lie}(V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Lie}(V\times V)$ is defined on
pure Lie brackets as follows:

$\partial_{V}\left(\left[v_{1},[v_{2},\ldots[v_{n-1},v_{n}]\ldots]\right]\right)=\sum^{n}_{i=1}\left[(v_{1},0),\left[(v_{2},0),\ldots\left[(0,v_{i}),\ldots[(v_{n-1},0),(v_{n},0)]\ldots\right]\right]\right].$

Therefore, $\partial$ corresponds to differentiating Lie bracket polynomials.

###  4.2 The Cartesian Differential Categories of an Operad

The first consequence of Theorem 4.1.1 is that the opposite category of the
Kleisli category of a monad associated to an operad is a Cartesian
differential category. As a shorthand, for an operad $\mathcal{P}$ we denote
$\mathsf{KL}_{\mathcal{P}}:=\mathsf{KL}_{\mathsf{S}(\mathcal{P},-)}$ for the
Kleisli category of $(\mathsf{S}(\mathcal{P},-),\gamma,\eta)$. So we may state
that:

######  Proposition 4.2.1.

Let $\mathcal{P}$ be an operad. Then $\mathsf{KL}_{\mathcal{P}}^{op}$, is a
Cartesian differential category.

Let us unpack this Cartesian differential category. Recall that the objects of
$\mathsf{KL}^{op}_{\mathcal{P}}$ are $R$-modules, while a map
$f:V\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}W$
in $\mathsf{KL}^{op}_{\mathcal{P}}$ is an $R$-linear morphism of type
$f:W\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V)$. The
derivative ${\mathsf{D}[f]:V\times V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}W}$ in $\mathsf{KL}^{op}_{\mathcal{P}}$
is the $R$-linear morphism of type
${\mathsf{D}[f]:W\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V\times V)}$
defined as:

$\mathsf{D}[f]=\partial_{V}\circ f.$

Let us give a bit of intuition about this. Let $X$ be a basis for a free
$R$-module $V$. Then $(\mu;x_{1},\ldots,x_{n})$ can be interpreted as a sort
of monomial of degree $n$, which we call $\mathcal{P}$-monomials. With this in
mind, an arbitrary element $P\in\mathsf{S}(\mathcal{P},V)$ is therefore a
finite sum of $\mathcal{P}$-monomials, and therefore we may interpret $P$ as a
$\mathcal{P}$-polynomial. Now
$(\mu;(x_{1},0),\ldots,(0,x_{i}),\ldots,(x_{n},0))$ should be understood as
the partial derivative of $(\mu;x_{1},\ldots,x_{n})$ in the variable $x_{i}$.
Therefore, the differential combinator transformation
$\partial_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V\times V)$ maps
$\mathcal{P}$-polynomials to the sum of their partial derivatives, which we
suggestively write as:

$\partial_{V}(P)=\sum_{x\in X}\frac{\partial P}{\partial x}dx,$

where the sum is well-defined since $P$ only depends on a finite number of
elements of $X$. In other words, $\partial_{V}$ maps $\mathcal{P}$-polynomials
to their total derivative. Now let us extend this intuition to the Kleisli
category. If $W$ is another free $R$-module with basis $Y$, then a map
$f:V\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}W$
in $\mathsf{KL}^{op}_{\mathcal{P}}$ is precisely associated to a tuple of
$\mathcal{P}$-polynomials in variables $X$:

$f\equiv\langle P_{y}\rangle_{y\in Y}.$

So its derivative $\mathsf{D}[f]:V\times V\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}W$ in $\mathsf{Kl}_{\mathcal{P}}^{op}$
is associated to the tuple of the total derivative of each
$\mathcal{P}$-polynomial:

$\mathsf{D}[f]\equiv\left\langle\sum_{x\in X}\frac{\partial P_{y}}{\partial
x}dx\right\rangle_{y\in Y}.$

Therefore, $\mathsf{KL}^{op}_{\mathcal{P}}$ can naively be understood as a
generalized Lawvere theory of $\mathcal{P}$-polynomial. We can obtain a
legitimate Lawvere theory of $\mathcal{P}$-polynomials by defining
$\mathcal{P}\text{-}\mathsf{POLY}$ to be the category whose objects are the
natural numbers $n\in\mathbb{N}$ and where a map
$n\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}m$
is an $m$-tuple $\langle P_{1},\ldots,P_{m}\rangle$ where
$P_{i}\in\mathsf{S}(\mathcal{P},R^{n})$. Then
$\mathcal{P}\text{-}\mathsf{POLY}$ is equivalent to the full subcategory of
$\mathsf{KL}^{op}_{\mathcal{P}}$ of finite dimensional $R$-modules, where in
particular
$\mathcal{P}\text{-}\mathsf{POLY}(n,m)\cong\mathsf{MOD}_{R}\left(R^{m},\mathsf{S}(\mathcal{P},R^{n})\right)$.
Also, note that
$\mathcal{P}\text{-}\mathsf{POLY}(n,1)=\mathsf{S}(\mathcal{P},R^{n})$.
Therefore we have that:

######  Proposition 4.2.2.

Let $\mathcal{P}$ be an operad. Then $\mathcal{P}\text{-}\mathsf{POLY}$ is a
Cartesian differential category where in particular for a map
$P:n\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}m$, $P=\langle
P_{1},\ldots,P_{m}\rangle$, its derivative $\mathsf{D}[n]:n\times
n\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}m$ is
defined as follows:

$\mathsf{D}\left[\langle
P_{1},\ldots,P_{m}\rangle\right]=\langle\partial_{R^{n}}(P_{1}),\ldots,\partial_{R^{n}}(P_{m})\rangle.$

Here is the resulting Cartesian differential category for our main examples of
operads.

######  Example 4.2.3.

For an $R$-algebra $A$, $A^{\bullet}\text{-}\mathsf{POLY}$ is equivalent to
the Cartesian differential category of $A$-linear maps. So let
$A\text{-}\mathsf{LIN}$ be the category whose objects are $n\in\mathbb{N}$ and
where a map $f:n\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}m$ is an $A$-linear morphism
$f:A^{n}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{m}$. Then $A\text{-}\mathsf{LIN}$ is
a Cartesian differential category where the differential combinator is defined
as $\mathsf{D}[f](\vec{x},\vec{y})=f(\vec{y})$ [22, Example 2.6], and
furthermore we have that $A^{\bullet}\text{-}\mathsf{POLY}\simeq
A\text{-}\mathsf{LIN}$ as Cartesian differential categories.

######  Example 4.2.4.

For the operad ${\operatorname{Com}}$,
${\operatorname{Com}}\text{-}\mathsf{POLY}$ recaptures precisely polynomial
differentiation since it is equivalent to the Lawvere theory of polynomials
over $R$, $R\text{-}\mathsf{POLY}$, which is one of the main examples of
Cartesian differential categories [22, Example 2.6]. Concretely,
$R\text{-}\mathsf{POLY}$ is the category whose objects are $n\in\mathbb{N}$
and where a map ${P:n\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}m}$ is a $m$-tuple of polynomials in $n$
variables, that is, $P=\langle p_{1}(\vec{x}),\ldots,p_{m}(\vec{x})\rangle$
with $p_{i}(\vec{x})\in R[x_{1},\ldots,x_{n}]$. $R\text{-}\mathsf{POLY}$ is a
Cartesian differential category where the differential combinator is given by
the standard differentiation of polynomials, that is, for a map
${P:n\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}m}$, with $P=\langle
p_{1}(\vec{x}),\ldots,p_{m}(\vec{x})\rangle$, its derivative
$\mathsf{D}[P]:n\times n\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}m$ is defined as the tuple of the sum of
the partial derivatives of the polynomials $p_{i}(\vec{x})$:

$\displaystyle\mathsf{D}[P](\vec{x},\vec{y}):=\left(\sum\limits^{n}_{i=1}\frac{\partial
p_{1}(\vec{x})}{\partial
x_{i}}y_{i},\ldots,\sum\limits^{n}_{i=1}\frac{\partial
p_{n}(\vec{x})}{\partial x_{i}}y_{i}\right),$
$\displaystyle\sum\limits^{n}_{i=1}\frac{\partial p_{j}(\vec{x})}{\partial
x_{i}}y_{i}\in R[x_{1},\ldots,x_{n},y_{1},\ldots,y_{n}].$

So we have that ${\operatorname{Com}}\text{-}\mathsf{POLY}\simeq
R\text{-}\mathsf{POLY}$ as Cartesian differential categories.

######  Example 4.2.5.

For the operad ${\operatorname{Ass}}$,
${\operatorname{Ass}}\text{-}\mathsf{POLY}$ captures instead differentiating
non-commutative polynomials since it is equivalent to the Lawvere theory of
non-commutative polynomials.

######  Example 4.2.6.

For the operad ${\operatorname{Lie}}$,
${\operatorname{Lie}}\text{-}\mathsf{POLY}$ is the cartesian differential
which is given by the Lawvere theory of Lie bracket polynomials.

We conclude this section by discussing the notion of a $\mathsf{D}$-linear
counit for a coCartesian differential monad.

######  Definition 4.2.7.

[22, Definition 3.8] For a coCartesian differential monad
$(\mathsf{S},\mu,\eta,\partial)$ on a semi-additive category $\mathbb{X}$, a
$\mathsf{D}$-linear counit is a natural transformation
$\mathcal{E}_{A}:\mathsf{S}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ in $\mathbb{X}$ such that the
following equalities hold:

[DU.1]

$\mathcal{E}_{A}\circ\eta_{A}=1_{A}$,

[DU.2]

$\eta_{A}\circ\mathcal{E}_{A}=\mathsf{S}(\pi_{2})\circ\partial_{A}$.

In a Cartesian differential category with a differential combinator
$\mathsf{D}$, there is an important class of maps called the
$\mathsf{D}$-linear maps [3, Definition 2.2.1] which are maps $f$ such that
$\mathsf{D}[f]=f\circ\pi_{2}$. For a coCartesian differential monad, the
$\mathsf{D}$-linear maps in the opposite category of its Kleisli category
correspond precisely to the maps in the base category if and only if the
coCartesian differential monad has a $\mathsf{D}$-linear counit [22,
Proposition 3.11]. We will now show that for an operad $\mathcal{P}$, its
associated monad has a $\mathsf{D}$-linear counit
$\mathcal{E}_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}V$ if and only if $\mathcal{P}(1)$ is of
dimension one (as an $R$-module). Essentially, note that
$\mathcal{P}(1)\otimes V\subset\mathsf{S}(\mathcal{P},V)$, and therefore, if
$\mathcal{P}(1)\cong R$, then there is a copy of $V$ inside
$\mathsf{S}(\mathcal{P},V)$. So the $\mathsf{D}$-linear counit amounts to
projecting out the $V$ component of $\mathsf{S}(\mathcal{P},V)$.

######  Lemma 4.2.8.

Let $\mathcal{P}$ be an operad. Then
$(\mathsf{S}(\mathcal{P},-),\gamma,\eta,\partial)$ has a $\mathsf{D}$-linear
counit
$\mathcal{E}_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}V$ if and only if the $R$-linear
morphism $e_{\mathcal{P}}:R\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}(1)$ which picks out the
distinguished element, $e_{\mathcal{P}}(1)=1_{\mathcal{P}}$, is an
isomorphism. Explicitly, if $e_{\mathcal{P}}$ is an isomorphism, define
$\mathcal{E}_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}V$ as follows:

$\displaystyle\mathcal{E}_{V}(\mu;v)=e^{-1}_{\mathcal{P}}(\mu)\cdot v\mbox{ if
}\mu\in\mathcal{P}(1),$
$\displaystyle\mathcal{E}_{V}(\mu;v_{1},\ldots,v_{n})=0\mbox{ if }n\neq 1,$

and conversely if
$\mathcal{E}_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}V$ is a $\mathsf{D}$-linear counit, then
define $e^{-1}_{\mathcal{P}}:\mathcal{P}(1)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}R$ as
$e^{-1}_{\mathcal{P}}(\mu)=\mathcal{E}_{R}(\mu;1)$.

###### Proof.

Suppose that $e_{\mathcal{P}}$ is an isomorphism. We must check that
$\mathcal{E}$ satisfies $\mathcal{E}_{V}\circ\eta_{V}=1_{V}$ and
$\eta_{V}\circ\mathcal{E}_{V}=\mathsf{S}(\mathcal{P},\pi_{2})\circ\partial_{V}$.
The first identity is automatic since
$e^{-1}_{\mathcal{P}}(1_{\mathcal{P}})=1$:

$\mathcal{E}_{V}(\eta_{v}(v))=\mathcal{E}_{V}(1_{\mathcal{P}};v)=e^{-1}_{\mathcal{P}}(1_{\mathcal{P}})\cdot
v=1\cdot v=v.$

For the other identity, we must prove it in two cases. For the case
$\mu\in\mathcal{P}(1)$, note that $\mu=e^{-1}_{\mathcal{P}}(\mu)\cdot
1_{\mathcal{P}}$. So, using that $(r\cdot\mu;v)=(\mu;r\cdot v)$, we have that:

$\displaystyle\mathsf{S}(\mathcal{P},\pi_{2})\left(\partial_{V}(\mu;v)\right)$
$\displaystyle=(\mu;v)=(e^{-1}_{\mathcal{P}}(\mu)\cdot
1_{\mathcal{P}};v)=(1_{\mathcal{P}};e^{-1}_{\mathcal{P}}(\mu)\cdot v)$
$\displaystyle=(1_{\mathcal{P}};\mathcal{E}_{V}(\mu,v))=\eta_{V}(\mathcal{E}_{V}(\mu,v)).$

Lastly, when $n\neq 1$, using that $(\mu;v_{1},\ldots,0,\ldots,v_{n})=0$, we
compute that:

$\displaystyle\mathsf{S}(\mathcal{P},\pi_{2})\left(\partial_{V}(\mu;v_{1},\ldots,v_{n})\right)=\sum^{n}_{i=1}(\mu;0,\ldots,v_{i},\ldots,0)=0=\eta_{V}(\mathcal{E}_{V}(\mu;v_{1},\ldots,v_{n})),$

so we have that $\mathcal{E}$ is a $\mathsf{D}$-linear counit. Conversely,
suppose that $\mathcal{E}$ is a $\mathsf{D}$-linear counit, so in particular,
$\mathcal{E}_{R}\circ\eta_{R}=1$ and
$\eta_{R}\circ\mathcal{E}_{R}=\mathsf{S}(\mathcal{P},\pi_{2})\circ\partial_{R}$.
On the one hand, we have:

$\displaystyle
e^{-1}_{\mathcal{P}}(e_{\mathcal{P}}(1))=\mathcal{E}_{R}(1_{\mathcal{P}};1)=\mathcal{E}_{R}(\eta_{R}(1))=1.$

On the other hand, first observe that by our notation, for
$\mu,\nu\in\mathcal{P}(1)$, if $(\mu;1)=(\nu;1)$, then this means that
$\mu=\nu$. We compute:

$\displaystyle(e_{\mathcal{P}}(e^{-1}_{\mathcal{P}}(\mu));1)=(\mathcal{E}_{R}(\mu;1)\cdot
1_{\mathcal{P}};1)=(1_{\mathcal{P}};\mathcal{E}_{R}(\mu;1)$
$\displaystyle=\eta_{R}(\mathcal{E}_{R}(\mu;1))=\mathsf{S}(\mathcal{P},\pi_{2})\left(\partial_{R}(\mu;1)\right)=(\mu;1),$

hence $e_{\mathcal{P}}(e^{-1}_{\mathcal{P}}(1)(\mu))=\mu$, and so,
$e_{\mathcal{P}}$ is an isomorphism. ∎

For the operads ${\operatorname{Com}}$, ${\operatorname{Ass}}$, and
${\operatorname{Lie}}$, their associated coCartesian differential monads all
have a $\mathsf{D}$-linear counit which is given precisely by projecting out
the copy of $V$ in the symmetric algebra, tensor algebra, or free Lie algebra
respectively. On the other hand, for an arbitrary $R$-algebras $A$, the
associated coCartesian differential monad of the operad $A^{\bullet}$ will
not, in general, have a $\mathsf{D}$-linear counit unless $A\cong R$.

###  4.3 Tangent Structure for Algebras over an Operad

In this section, we describe the tangent structure on the category of algebras
over an operad. We have already shown that the monad associated to an operad
is a coCartesian differential monad, then, by applying Proposition 3.2.4 and
Theorem 3.2.7, we can state the following:

######  Lemma 4.3.1.

Let $\mathcal{P}$ be an operad. For the coCartesian differential monad
$(\mathsf{S}(\mathcal{P},-),\gamma,\eta,\partial)$, the induced
$\mathbb{B}$-distributive law $\lambda_{V}:\mathsf{S}(\mathcal{P},V\times
V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V)\times\mathsf{S}(\mathcal{P},V)$
as defined in Proposition 3.2.4 is given as follows:

$\lambda_{V}(\mu;(u_{1},v_{1}),\dots,(u_{n},v_{n}))=\left((\mu;u_{1},\dots,u_{n}),\sum_{i=1}^{n}\left(\mu;u_{1},\ldots,v_{i},\ldots,u_{n}\right)\right).$

Furthermore, $(\mathsf{ALG}_{\mathsf{S}(\mathcal{P},-)},\mathbb{T})$ is a
Cartesian Rosický tangent category, where the Rosický tangent structure is
defined as in Theorem 3.2.7.

###### Proof.

Recall that, by definition,
$\lambda_{V}:=\left\langle\mathsf{S}(\mathcal{P},\pi_{1}),S\left(\mathcal{P},\pi_{1}+\pi_{4}\right)\circ\partial_{V\times
V}\right\rangle$. We leave it as an exercise for the reader to check that this
gives precisely the formula above. ∎

Let us give a more concrete description of the tangent structure by describing
the tangent bundle in terms of semi-direct products. Following the standard
terminology in operad literature, by an algebra over an operad we mean an
algebra over the operad’s associated monad [28, Section 5.2.3]. More
explicitly, for an operad $\mathcal{P}$, a $\mathcal{P}$-algebra is an
$R$-module $A$ equipped with an $R$-linear morphism
$\theta:\mathsf{S}(\mathcal{P},A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ such that $\theta\circ\eta_{A}=1_{A}$
and $\theta\circ\mathsf{S}(\mathcal{P},\theta)=\theta\circ\gamma_{A}$. As a
useful shorthand, we write:

$\displaystyle\theta(\mu;a_{1},\dots,a_{n}):=\mu(a_{1},\dots,a_{n}),$

and so, we will only write $A$ for a $\mathcal{P}$-algebra when there is no
confusion about its $\mathcal{P}$-algebra structure $\theta$. Therefore, the
necessary $\mathcal{P}$-algebra identities can be expressed as:

$\displaystyle 1_{\mathcal{P}}(a)=a,$
$\displaystyle\mu\left(\nu_{1}(a_{1,1},\ldots,a_{1,n_{1}}),\ldots,\nu_{k}(v_{k,1},\ldots,v_{k,n_{k}})\right)=\mu\left(\nu_{1},\ldots,\nu_{k}\right)(a_{1,1},\ldots,a_{k,n_{k}}).$

Similarly, by a $\mathcal{P}$-algebra morphism we simply mean an
$\mathsf{S}(\mathcal{P},-)$-algebra morphism. So if $A$ and $B$ are
$\mathcal{P}$-algebras, then a $\mathcal{P}$-algebra morphism is an $R$-linear
morphism $f:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B$ which is compatible with
$\mathcal{P}$-algebra structure, that is, the following equality holds:

$\displaystyle f(\mu(a_{1},\dots,a_{n}))=\mu(f(a_{1}),\dots,f(a_{n})).$

Therefore, by the category of algebras over an operad we mean the
Eilenberg–Moore category of its associated monad. So let
$\mathsf{ALG}_{\mathcal{P}}$ be the category of $\mathcal{P}$-algebras and
$\mathcal{P}$-algebra morphisms between them, or in other words,
$\mathsf{ALG}_{\mathcal{P}}:=\mathsf{ALG}_{\mathsf{S}(\mathcal{P},-)}$.

By Lemma 4.3.1, we already know that $\mathsf{ALG}_{\mathcal{P}}$ is a tangent
category. However, we wish to give a more explicit description of the tangent
bundle of a $\mathcal{P}$-algebra. We do this using the semi-direct product of
a $\mathcal{P}$-algebra with itself. We note that, while the semi-direct
product [28, Section 12.3.2] can be more generally defined between a
$\mathcal{P}$-algebra and a module (a notion that we review in the next
section below), for the purpose of the tangent structure, we only need to
understand it for a $\mathcal{P}$-algebra with itself.

For a $\mathcal{P}$-algebra $A$, define the $\mathcal{P}$-algebra $A\ltimes A$
as the $R$-module $A\times A$ equipped with $\mathcal{P}$-algebra structure
defined as follows:

$\displaystyle\mu((a_{1},b_{1}),\ldots,(a_{n},b_{n}))=\left(\mu(a_{1},\ldots,a_{n}),\sum_{i=1}^{n}\mu(a_{1},\ldots,b_{i},\ldots,a_{n})\right).$

Note that, in $A\ltimes A$, the first component $A$ is viewed as a
$\mathcal{P}$-algebra, while the second component $A$ is viewed as a module.
More generally, we also define the $\mathcal{P}$-algebra $A\ltimes A^{n}$ to
be the $R$-module $A\times\underbrace{A\times\ldots\times A}_{n\text{ times}}$
with $\mathcal{P}$-algebra structure defined as (where we denoted by $\vec{b}$
a tuple $\left(b_{1},\ldots,b_{n}\right)$ of elements of $A$):

$\mu\left((a_{1},\vec{b}_{1}),\ldots,(a_{m},\vec{b}_{m})\right)=\\\
\left(\mu(a_{1},\ldots,a_{m}),\sum_{i=1}^{m}\mu(a_{1},\ldots,b_{i,1},\ldots,a_{m}),\ldots,\sum_{i=1}^{m}\mu(a_{1},\ldots,b_{i,n},\ldots,a_{m})\right).$

We will now prove that $A\ltimes A$ is precisely the tangent bundle over $A$.

######  Lemma 4.3.2.

Let $\mathcal{P}$ be an operad and $A$ a $\mathcal{P}$-algebra. Then
$\mathsf{T}(A)=A\ltimes A$ and $\mathsf{T}_{n}(A)=A\ltimes A^{n}$, where
$\mathsf{T}(A)$ and $\mathsf{T}_{n}(A)$ are defined as in Theorem 3.2.7.

###### Proof.

Let $\theta:\mathsf{S}(\mathcal{P},A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ denote the $\mathcal{P}$-algebra
structure map of $A$. Then, by Theorem 3.2.7, $\mathsf{T}(A)$ is the
$R$-module $A\times A$ with $\mathcal{P}$-algebra structure map defined as
$(\theta\times\theta)\circ\lambda_{A}$, where $\lambda$ was given in Lemma
4.3.1. So, since $\mathsf{T}(A)$ and $A\ltimes A$ have the same underlying
$R$-module $A\times A$, we must show that they have the same
$\mathcal{P}$-algebra structure. We compute:

$\displaystyle(\theta\times\theta)\left(\lambda_{A}\left((a_{1},b_{1}),\dots,(a_{n},b_{n})\right)\right)$
$\displaystyle=(\theta\times\theta)\left((\mu;a_{1},\dots,a_{n}),\sum_{i=1}^{n}\left(\mu;a_{1},\ldots,b_{i},\ldots,a_{n}\right)\right)$
$\displaystyle=\left(\mu(a_{1},\ldots,a_{n}),\sum_{i=1}^{n}\mu\left(a_{1},\ldots,b_{i},\ldots,a_{n}\right)\right).$

We conclude that $\mathsf{T}(A)=A\ltimes A$. Similarly, we can also show that
$\mathsf{T}_{n}(A)=A\ltimes A^{n}$. ∎

Therefore, we may write the tangent structure of Theorem 3.2.7 for
$\mathsf{ALG}_{\mathcal{P}}$ in terms of semi-direct products.

######  Theorem 4.3.3.

Let $\mathcal{P}$ be an operad. Consider:

  1. (i)

The tangent bundle functor
$\mathsf{T}:\mathsf{ALG}_{\mathcal{P}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}$ defined on
objects as $\mathsf{T}(A)=A\ltimes A$ and on maps as $\mathsf{T}(f)=(f\times
f)$, that is:

$\mathsf{T}(f)(a,b)=(f(a),f(b))$

  2. (ii)

The projection $p_{A}:A\ltimes A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ defined as:

$p_{A}(a,b)=a$

with $n$-fold pullback $\mathsf{T}_{n}(A)=A\ltimes A^{n}$, and projections
$q_{j}:A\ltimes A^{n}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\ltimes A$ defined as:

$q_{j}(a,b_{1},\ldots,b_{n})=(a,b_{j})$

  3. (iii)

The sum $s_{A}:A\ltimes A^{2}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\ltimes A$ defined as:

$s_{A}(a,b_{1},b_{2})=(a,b_{1}+b_{2})$

  4. (iv)

The zero map $z_{A}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\ltimes A$ defined as:

$z_{A}(a)=(a,0)$

  5. (v)

The vertical lift $\ell_{A}:A\ltimes A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(A\ltimes A)\ltimes(A\ltimes A)$ defined
as:

$\ell_{A}(a,b)=(a,0,0,b)$

  6. (vi)

The canonical flip $c_{A}:(A\ltimes A)\ltimes(A\ltimes
A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(A\ltimes A)\ltimes(A\ltimes A)$ defined
as:

$c_{A}(a,b,c,d)=(a,c,b,d)$

  7. (vii)

The negative map $n_{A}:A\ltimes A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\ltimes A$ defined as:

$n_{A}(a,b)=(a,-b)$

Then, $\mathbb{T}=(\mathsf{T},p,s,z,l,c,n)$ is a Rosický tangent structure on
$\mathsf{ALG}_{\mathcal{P}}$, and so,
$(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ is a Cartesian Rosický tangent
category.

We now consider what the resulting tangent categories are for our main
examples of operads.

######  Example 4.3.4.

Let $A$ be an $R$-algebra. Then for the operad $A^{\bullet}$, the
$A^{\bullet}$-algebras are precisely $A$-modules. So,
$\mathsf{ALG}_{A^{\bullet}}=\mathsf{MOD}_{A}$, the category of $A$-modules and
$A$-linear maps between them. The resulting tangent structure on
$\mathsf{MOD}_{A}$ is precisely the biproduct structure from Lemma 3.1.1, so
in particular, for an $A$-module $M$, its tangent bundle is simply
$\mathsf{T}(M)=M\times M$. So $(\mathsf{MOD}_{A},\mathbb{B})$ is a Cartesian
Rosický tangent category.

######  Example 4.3.5.

For the operad ${\operatorname{Com}}$, the ${\operatorname{Com}}$-algebras are
precisely (associative and unital) commutative $R$-algebras. So
$\mathsf{ALG}_{\operatorname{Com}}=\mathsf{CALG}_{R}$, the category of
commutative $R$-algebras and $R$-algebra morphisms between them. Up to
isomorphism, the resulting tangent structure is the one described in [12,
Section 2.2], where the tangent bundle is given by dual numbers. Indeed, for a
commutative $R$-algebra $A$, let $A[\epsilon]$ be its $R$-algebra of dual
numbers, which we use as a shorthand for $A[\epsilon]:=A[x]/(x^{2})$:

$\displaystyle A[\epsilon]=\\{a+b\epsilon|~{}\forall a,b\in A\\},$
$\displaystyle\text{with }\epsilon^{2}=0.$

It is easy to see that $A\ltimes A\cong A[\epsilon]$ via $(a,b)\mapsto
a+b\epsilon$. So we may express the tangent structure using instead dual
numbers, so $\mathsf{T}(A)=A[\epsilon]$. We may write $\mathsf{T}^{2}(A)$ and
$\mathsf{T}_{n}(A)$ as multivariable dual numbers in the following way:

$\displaystyle\mathsf{T}^{2}(A)=A[\epsilon][\epsilon^{\prime}]=\\{a+b\epsilon+c\epsilon^{\prime}+d\epsilon\epsilon^{\prime}|~{}\forall
a,b,c,d\in A\\},$ $\displaystyle\text{with
}\epsilon^{2}={\epsilon^{\prime}}^{2}=0,$
$\displaystyle\mathsf{T}_{n}(A)=A[\epsilon_{1},\ldots,\epsilon_{n}]=\\{a+b_{1}\epsilon_{1}+\ldots+b_{n}\epsilon_{n}|~{}\forall
a,b_{j}\in A\\},$ $\displaystyle\text{with }\epsilon_{i}\epsilon_{j}=0.$

The rest of the tangent structure is given as follows:

$\displaystyle p_{A}(a+b\epsilon)=a,\quad\quad\quad
s_{A}(a+b\epsilon_{1}+c\epsilon_{2})=a+(b+c)\epsilon,\quad\quad\quad
z_{A}(a)=a,$ $\displaystyle
l_{A}(a+b\epsilon)=a+b\epsilon\epsilon^{\prime},\quad\quad\quad
c_{A}(a+b\epsilon+c\epsilon^{\prime}+d\epsilon\epsilon^{\prime})=a+c\epsilon+b\epsilon^{\prime}+d\epsilon\epsilon^{\prime},$
$\displaystyle n_{A}(a+b\epsilon)=a-b\epsilon,$

so $(\mathsf{CALG}_{R},\mathbb{T})$ is a Cartesian Rosický tangent category.

######  Example 4.3.6.

For the operad ${\operatorname{Ass}}$, the ${\operatorname{Ass}}$-algebras are
precisely (associative and unital) $R$-algebras. So
$\mathsf{ALG}_{\operatorname{Ass}}=\mathsf{ALG}_{R}$, the category of
$R$-algebras and $R$-algebra morphisms between them. Again, up to isomorphism,
the resulting tangent structure is precisely the same as for commutative
algebras in Example 4.3.5, so in particular for an $R$-algebra $A$,
$\mathsf{T}(A)=A[\epsilon]\cong A\ltimes A$. Therefore,
$(\mathsf{ALG}_{R},\mathbb{T})$ is a Cartesian Rosický tangent category.

######  Example 4.3.7.

For the operad ${\operatorname{Lie}}$, the ${\operatorname{Lie}}$-algebras are
precisely Lie algebras over $R$. So
$\mathsf{ALG}_{\operatorname{Ass}}=\mathsf{LIE}_{R}$, the category of Lie
algebras over $R$ and Lie algebra morphisms between them. For a Lie algebra
${\mathfrak{g}}$, the Lie algebra ${\mathfrak{g}}\ltimes{\mathfrak{g}}$ is the
$R$-module ${\mathfrak{g}}\times{\mathfrak{g}}$ with the Lie brackets defined
as
$\left[(x_{1},y_{1}),(x_{2},y_{2})\right]:=([x_{1},x_{2}],[x_{1},y_{2}]+[y_{1},x_{2}])$.
So $(\mathsf{LIE}_{R},\mathbb{T})$ is a Cartesian Rosický tangent category,
which we stress is a new example of a tangent category.

We conclude this section by mentioning that the construction of Theorem 4.3.3
provides a (contravariant) functor between the category of operads and the
category of Cartesian Rosický tangent categories. Briefly, for operads
$\mathcal{P}$ and $\mathcal{P}^{\prime}$, an operad morphism
$f:\mathcal{P}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}$ is a sequence of
equivariant $R$-linear morphisms
$f(n):\mathcal{P}(n)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}(n)$ which preserve
the partial compositions and the distinguished object. Then, let
$\mathsf{OPERAD}_{R}$ be the category of operads and operad morphisms between
them. On the other hand, for $(\mathbb{X},\mathbb{T})$ and
$(\mathbb{X}^{\prime},\mathbb{T}^{\prime})$, a strict Cartesian Rosický
tangent morphism
$\mathsf{F}:(\mathbb{X},\mathbb{T})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(\mathbb{X}^{\prime},\mathbb{T}^{\prime})$
is a functor $\mathsf{F}:\mathbb{X}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathbb{X}^{\prime}$ which preserves the
product strictly and also preserve the tangent structure, in the sense that
$\mathsf{F}\circ\mathsf{T}=\mathsf{T}^{\prime}\circ\mathsf{F}$,
$\mathsf{F}(p)=p^{\prime}$, etc. Let $\mathsf{CRTAN}_{=}$ be the category of
Cartesian Rosický tangent categories and strict Cartesian tangent morphisms
between them. Every operad morphism
${f:\mathcal{P}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}}$ induces a functor
$\mathsf{ALG}_{f}:\mathsf{ALG}_{\mathcal{P}^{\prime}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}$ by mapping a
$\mathcal{P}^{\prime}$-algebra $A$ to $A$ with $\mathcal{P}$-algebra structure
defined as $\mu(a_{1},\ldots,a_{n}):=f(\mu)(a_{1},\ldots,a_{n})$. It is
straightforward to see that
$\mathsf{ALG}_{f}:(\mathsf{ALG}_{\mathcal{P}^{\prime}},\mathbb{T})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$
is a strict Cartesian Rosický tangent morphism. Therefore we obtain a functor
$\mathsf{ALG}:\mathsf{OPERAD}^{op}_{R}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{CRTAN}_{=}$ which sends an
operad $\mathcal{P}$ to $(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ and operad
morphisms $f:\mathcal{P}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}$ to
$\mathsf{ALG}_{f}:(\mathsf{ALG}_{\mathcal{P}^{\prime}},\mathbb{T})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$.

In future work, it will be interesting to see if the left adjoint of the
pullback functor $\mathsf{ALG}_{f}$ induced by an operad morphism $f$ is also
a tangent morphism.

###  4.4 Adjoint Tangent Structure of Algebras over an Operad

The objective of this section is to study a tangent structure for the opposite
category of algebras over an operad, which is adjoint to the one we defined in
Section 4.3.

Since the category of algebras over an operad is always cocomplete, it admits
all reflexive coequalizers. So, by Theorem 3.3.2, we deduce that the tangent
structure defined in Section 4.3 has adjoint tangent structure. The adjoint
tangent bundle is given by the free algebra over the module of Kähler
differentials of an algebra. This is quite a mouthful, so let’s break it down
piece by piece.

Let $\mathcal{P}$ be an operad and $A$ a $\mathcal{P}$-algebra. Then an
$A$-module [28, Section 12.3.1] is an $R$-module $M$ equipped with a family of
$R$-linear morphisms $\psi_{n+1}:\mathcal{P}(n+1)\otimes A^{\otimes n}\otimes
M\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}M$,
called evaluation maps, satisfying natural equivariance, associativity, and a
unit map: $\eta_{M}:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\bigoplus_{n\in\mathbb{N}}\mathcal{P}(n+1)\otimes
A^{\otimes n}\otimes M$ playing the role of a unit for the evaluation. As a
shorthand, we write:

$\displaystyle\mu(a_{1},\dots,a_{n},x):=\psi_{n+1}(\mu\otimes
a_{1}\otimes\dots\otimes a_{n}\otimes x).$

If $M$ and $M^{\prime}$ are $A$-modules, then an $A$-linear morphism is an
$R$-linear morphism $f:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M^{\prime}$ which preserves the
evaluation and unit maps in the sense that:

$\displaystyle\eta_{M\prime}(f(x))$
$\displaystyle=\left(\bigoplus_{n+1}\mathcal{P}(n+1)\otimes A^{\otimes
n}\otimes f\right)\circ\eta_{M}(x),$ $\displaystyle
f(\mu(a_{1},\dots,a_{n},x))$ $\displaystyle=\mu(a_{1},\dots,a_{n},f(x)).$

Let $\mathsf{MOD}_{A}$ be the category of $A$-modules and $A$-linear morphisms
between them. Among the $A$-modules, there is an important one called the
module of Kähler differentials of $A$, which generalizes the classical notion
of Kähler differentials.

We must first describe derivations for algebras over an operad. For an
$A$-module $M$, we adopt the following useful notation:

$\displaystyle\mu(a_{1},\dots,a_{i},x,a_{i+1},\dots,a_{n})=\left(\mu\cdot(i~{}i+1\ldots
n)\right)(a_{1},\ldots,a_{n},x),$

where $(i~{}i+1\dots n)$ is the $n+1-i$-cycle permutation. An $A$-derivation
[28, Section 12.3.7] evaluated in an $A$-module $M$ is an $R$-linear morphism
$D:A\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}M$
satisfying:

$\displaystyle
D(\mu(a_{1},\dots,a_{n}))=\sum_{i=1}^{n}\mu(a_{1},\ldots,D(a_{i}),\ldots,a_{n}).$

This equality is called the Leibniz rule.

Now let ${\operatorname{Der}}(A,M)$ be the $R$-module of $A$-derivations
evaluated in $M$. This way, we define a functor ${\operatorname{Der}}(A,-)$
which is representable [28, Proposition 12.3.11]. The module of Kähler
differentials of $A$ [28, Section 12.3.8] is an $A$-module which represents
${\operatorname{Der}}(A,-)$, that is, an $A$-module $\Omega_{A}$ such that,
for all $A$-modules $M$,
${\operatorname{Der}}(A,M)\cong\mathsf{MOD}_{A}(\Omega_{A},M)$. This means
that there is an $A$-derivation $\mathsf{d}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\Omega_{A}$ which is universal in the
sense that for every $A$-derivation ${D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M}$, there exists a unique $A$-linear
morphism $\overline{D}:\Omega_{A}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M$ such that
$\overline{D}\circ\mathsf{d}=D$. We do not need a concrete description of
$\Omega_{A}$ for this paper, see [28, Lemma 12.3.12] for full details.
However, it is interesting to point out that, for the $\mathcal{P}$-algebra
$\mathsf{S}(\mathcal{P},V)$, $\Omega_{\mathsf{S}(\mathcal{P},V)}$ is
isomorphic to the sub-$\mathsf{S}(\mathcal{P},V)$-module of
$\mathsf{S}(\mathcal{P},V\times V)$ generated by elements of the form
$\left(\mu;(v_{1},0),\ldots,(0,v_{i}),\ldots,(v_{n},0)\right)$. Therefore, the
differential combinator transformation
$\partial_{V}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V\times V)$
factors through $\Omega_{\mathsf{S}(V)}$ by composing the derivation
$\mathsf{d}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\Omega_{\mathsf{S}(V)}$ with the
inclusion $\Omega_{\mathsf{S}(V)}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V\times V)$.

For an arbitrary $\mathcal{P}$-algebra $A$, $\Omega_{A}$ is not a
$\mathcal{P}$-algebra. One might be tempted to consider
$\mathsf{S}(\mathcal{P},\Omega_{A})$ as a candidate for the tangent bundle
over $A$. However, this is not the adjoint functor we are looking for. We will
instead take the free $A$-algebra over $\Omega_{A}$. An $A$-algebra, also
called a $\mathcal{P}$-algebra under $A$ [2], is a $\mathcal{P}$-algebra $B$
equipped with a $\mathcal{P}$-algebra morphism
$u:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B$. Now, if $B$ and $B^{\prime}$ are
$A$-algebras with $u:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B$ and
$u^{\prime}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B^{\prime}$ respectively, then an
$A$-algebra morphism is a $\mathcal{P}$-algebra morphism
$f:B\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B^{\prime}$ which also preserves the
$A$-algebra structure, so $f\circ u=u^{\prime}$. Let $\mathsf{ALG}_{A}$ be the
category of $A$-algebras and $A$-algebra morphisms between them. Every
$A$-algebra $B$ is also an $A$-module where the evaluation is given by:

$\displaystyle\mu(a_{1},\dots,a_{n},b):=\mu(u(a_{1}),\dots,u(a_{n}),b),$

and similarly, every $A$-algebra morphism is also an $A$-module morphism. We
obtain a functor
$\mathsf{U}_{A}:\mathsf{ALG}_{A}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{MOD}_{A}$. The functor
$\mathsf{U}_{A}$ has a left adjoint
$\mathsf{Free}_{A}:\mathsf{MOD}_{A}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{A}$. In the next
proposition, we provide a concrete description of $\mathsf{Free}_{A}$. This is
an extension of a result due to Ginzburg, who proved the existence of
$\mathsf{Free}_{A}$ for quadratic operads [17, Lemma 5.2]. This means that,
for every $A$-module $M$, there exists an $A$-algebra $\mathsf{Free}_{A}(M)$
with $u_{M}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{A}(M)$ called the free
$A$-algebra over $M$.

######  Proposition 4.4.1.

Let $A$ be a $\mathcal{P}$-algebra and let $M$ be a module over $A$ (in the
operadic sense). Consider the $\mathcal{P}$-algebra $\mathsf{Free}_{A}M$
obtained by quotienting the free algebra $\mathsf{S}(\mathcal{P},A\times M)$
by the ideal generated by the following relations:

$(\mu;(a_{1},0),\dots,(a_{k-1},0),(a_{k},x),(a_{k+1},0)\dots,(a_{n},0))\\\
=(\mu(a_{1},\dots,a_{n}),\mu(a_{1},\dots,a_{k-1},x,a_{k+1},\dots,a_{n})),$

for every $\mu\in\mathcal{P}(n)$, $a_{1},\dots,a_{n}\in A$, $x\in M$ and
positive integer $n$. Then,
$\mathsf{Free}_{A}:\mathsf{MOD}_{A}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{A}$ extends to a left
adjoint to the functor
$\mathsf{U}_{A}:\mathsf{ALG}_{A}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{MOD}_{A}$, where the $A$-algebra
structure $u_{M}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{A}M$ is defined as the
injection $u_{M}(a)=(a,0)$.

###### Proof.

Note that we have an $A$-module morphism
$\iota_{M}:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{A}M$ defined as the
inclusion $\iota_{M}(x)=(0,x)$. Now given an $A$-algebra
$u:A\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}B$
and an $A$-algebra morphism
$f:\mathsf{Free}_{A}M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B$, we can define an $A$-module morphsim
$f^{\flat}:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{A}M$ as the composite
$f^{\flat}=f\circ\iota_{M}$. Conversely, given an $A$-module morphism
$g:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{U}_{A}B$, it is not difficult to
check that the $\mathcal{P}$-algebra morphism $\mathsf{S}(\mathcal{P},A\times
M)\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}B$
mapping $(a,x)\mapsto u(a)+g(x)$ lifts to the quotient. As such, it provides a
well-defined $A$-algebra morphism
$g^{\sharp}:\mathsf{Free}_{A}M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}B$. The final step is to note that
$f\mapsto f^{\flat}$ and $g\mapsto g^{\sharp}$ are inverses of each other.
Thus, we obtain a natural bijection
$\mathsf{ALG}_{A}(\mathsf{Free}_{A}M,B)\cong\mathsf{MOD}_{A}(M,\mathsf{U}_{A}B)$,
and thus an adjunction as desired. ∎

With all this setup, we can finally define the adjoint tangent bundle of a
$\mathcal{P}$-algebra to be
$\mathsf{T}^{\circ}(A):=\mathsf{Free}_{A}(\Omega_{A})$. Using the combined
universal properties of both $\Omega_{A}$ and $\mathsf{Free}_{A}(-)$, we can
conclude that:

$\displaystyle\mathsf{ALG}_{\mathcal{P}}\left(\mathsf{Free}_{A}(\Omega_{A}),A^{\prime}\right)\cong\mathsf{ALG}_{\mathcal{P}}(A,A^{\prime}\ltimes
A^{\prime}).$

Therefore,
$\mathsf{T}^{\circ}:\mathsf{ALG}_{\mathcal{P}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}$ is indeed a
left adjoint to
$\mathsf{T}:\mathsf{ALG}_{\mathcal{P}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}$. However,
let us give a more concrete description of the adjoint tangent bundle and the
adjunction. For a $\mathcal{P}$-algebra $A$, its adjoint tangent bundle
$\mathsf{T}^{\circ}(A)$ is explicitly given by the $\mathcal{P}$-algebra
obtained by quotienting $\mathsf{S}(\mathcal{P},A\times A)$ by the following
relations:

$\displaystyle\mu((a_{1},0),\dots,(a_{n},0))$
$\displaystyle=(\mu(a_{1},\dots,a_{n}),0),$
$\displaystyle(0,\mu(a_{1},\dots,a_{n}))$
$\displaystyle=\sum_{i=1}^{n}\mu((a_{1},0),\ldots,(0,a_{i}),\ldots,(a_{n},0)).$

As useful shorthand, we write $a:=(a,0)\in\mathsf{T}^{\circ}(A)$ and
$\mathsf{d}(a):=(0,a)\in\mathsf{T}^{\circ}(A)$ for all $a\in A$. The above
relations then state that $\mu(a_{1},\dots,a_{n})$ in $\mathsf{T}^{\circ}(A)$
corresponds to $\mu(a_{1},\dots,a_{n})$ in $A$, and that

$\displaystyle\mathsf{d}\left(\mu(a_{1},\dots,a_{n})\right)$
$\displaystyle=\sum_{i=1}^{n}\mu(a_{1},\ldots,\mathsf{d}(a_{i}),\dots,a_{n}).$

Note that $\mathsf{d}$ can be thought of an $R$-linear function:
$\mathsf{d}(r\cdot a+s\cdot b)=r\cdot\mathsf{d}(a)+s\cdot\mathsf{d}(b)$. As a
$\mathcal{P}$-algebra, $\mathsf{T}^{\circ}(A)$ is generated by $a$ and
$\mathsf{d}(a)$ for all $a\in A$. As such, to define $\mathcal{P}$-algebra
morphisms with domain $\mathsf{T}^{\circ}(A)$, it suffices to define them on
the generators $a$ and $\mathsf{d}(a)$, making sure that the definition is
compatible with the above relations. For every $\mathcal{P}$-algebra morphism
$f:P\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}P^{\prime}$, define the
$\mathcal{P}$-algebra morphism
${\mathsf{T}^{\circ}(f):\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A^{\prime})}$ on
generators as follows:

$\displaystyle\mathsf{T}^{\circ}(f)(a)=f(a),$
$\displaystyle\mathsf{T}^{\circ}(f)(\mathsf{d}(a))=\mathsf{d}(f(a)).$

This gives us the desired functor
${\mathsf{T}^{\circ}:\mathsf{ALG}_{\mathcal{P}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}}$. Observe
that we did not need the $A$-algebra structure of $\mathsf{T}^{\circ}(A)$ to
build this functor. Nevertheless, readers familiar with modules of Kähler
differentials may easily check that the presentation of
$\mathsf{T}^{\circ}(A)$ given here recaptures precisely
$\mathsf{Free}_{A}(\Omega_{A})$ (especially using the $\mathsf{d}$ notation).
That said, the $A$-algebra structure of $\mathsf{T}^{\circ}(A)$ will be
precisely the adjoint projection
$p^{\circ}_{A}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A)$.

Turning our attention back to the adjunction, we define the unit
$\eta_{A}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A)\ltimes\mathsf{T}^{\circ}(A)$
as follows:

$\displaystyle\eta_{A}(a)=(a,\mathsf{d}(a)),$

which is clearly a $\mathcal{P}$-algebra morphism. The counit
$\varepsilon_{A}:\mathsf{T}^{\circ}(A\ltimes
A)\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}A$
is the $\mathcal{P}$-algebra morphism defined on generators $(a,b)$ and
$\mathsf{d}(a,b)$ for all $(a,b)\in A\ltimes A$ as follows:

$\displaystyle\varepsilon_{A}(a,b)=a,$
$\displaystyle\varepsilon(\mathsf{d}(a,b))=b.$

######  Lemma 4.4.2.

$(\eta,\varepsilon):\mathsf{T}^{\circ}\dashv\mathsf{T}$ is an adjunction.

###### Proof.

We leave it as an exercise for the reader to check that the adjunction
triangle identities are satisfied. Alternatively, one could check that
$\mathsf{ALG}_{\mathcal{P}}\left(\mathsf{T}^{\circ}(A),A^{\prime}\right)\cong\mathsf{ALG}_{\mathcal{P}}(A,A^{\prime}\ltimes
A^{\prime})$. Explicitly, given a $\mathcal{P}$-algebra morphism
$f:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$, define the
$\mathcal{P}$-algebra morphism $f^{\flat}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}\ltimes A^{\prime}$ as
$f^{\flat}(a)=(f(a),\mathsf{d}(f(a)))$, and conversely, given a
$\mathcal{P}$-algebra morphism $g:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}\ltimes A^{\prime}$, with
$g(a)=(g_{1}(a),g_{2}(a))$, define the $\mathcal{P}$-algebra morphism
$g^{\sharp}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$ on generators as
$g^{\sharp}(a)=g_{1}(a)$ and $g^{\sharp}(\mathsf{d}(a))=g_{2}(a)$. One can
check that the mappings $f\mapsto f^{\flat}$ and $g\mapsto g^{\sharp}$ define
mutually inverse bijections. ∎

For any operad $\mathcal{P}$, $\mathsf{ALG}_{\mathcal{P}}$ is cocomplete [30,
Proposition 6.4], and so, it admits all coproducts and pushouts. Therefore,
applying Theorem 2.2.2 and Corollary 2.2.4, we obtain:

######  Corollary 4.4.3.

Let $\mathcal{P}$ be an operad and $A$ a $\mathcal{P}$-algebra. The Cartesian
Rosický tangent category $(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ defined in
Theorem 4.3.3 has an adjoint tangent structure, and so
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ is a Cartesian Rosický
tangent category, where $\mathbb{T}^{\circ}$ is defined as in Theorem 2.2.2.

We will now give a concrete description of the adjoint tangent structure. We
can define all the necessary structure maps on generators. Let us first
describe the generators of $\mathsf{T}^{\circ}_{n}(A)$ and
${\mathsf{T}^{\circ}}^{2}(A)$. On the one hand, $\mathsf{T}^{\circ}_{n}(A)$ is
a quotient of $\mathsf{S}(\mathcal{P},\prod\limits^{n+1}_{i=1}A)$ modulo the
similar equations as above, and therefore, can be described in terms of
generators $a$ and $\mathsf{d}_{i}(a)$ for all $a\in A$ and $1\leq i\leq n$.
By Lemma 2.2.3, $\mathsf{T}^{\circ}_{n}$ is indeed a left adjoint to
$\mathsf{T}_{n}$. On the other hand, ${\mathsf{T}^{\circ}}^{2}(A)$ is
generated by $a$, $\mathsf{d}(a)$, $\mathsf{d}^{\prime}(a)$, and
$\mathsf{d}^{\prime}\mathsf{d}(a)$ for all $a\in A$, where $\mathsf{d}$ is for
$\Omega_{A}$ and $\mathsf{d}^{\prime}$ is for
$\Omega_{\mathsf{T}^{\circ}(A)}$.

######  Theorem 4.4.4.

Let $\mathcal{P}$ be an operad. Consider:

  1. (i)

The adjoint tangent bundle functor
$\mathsf{T}^{\circ}:\mathsf{ALG}_{\mathcal{P}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}_{\mathcal{P}}$ defined on
objects by $\mathsf{T}^{\circ}(A)=\mathsf{Free}_{A}(\Omega_{A})$ and on maps
by $\mathsf{T}^{\circ}(f)$ (as defined above);

  2. (ii)

The adjoint projection $p^{\circ}_{A}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A)$ defines as:

$p^{\circ}_{A}(a)=a$

and where the $n$-fold pushout of $p^{\circ}_{A}$ is
$\mathsf{T}^{\circ}_{n}(A)$, with injections
$q^{\circ}_{j}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}_{n}(A)$ defined on
generators as:

$\displaystyle q^{\circ}_{j}(a)=a$ $\displaystyle
q^{\circ}_{j}(\mathsf{d}(a))=\mathsf{d}_{j}(a)$

  3. (iii)

The adjoint sum
$s^{\circ}_{A}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}_{2}(A)$, defined on
generators as:

$\displaystyle s^{\circ}_{A}(a)=a$ $\displaystyle
s^{\circ}_{A}(\mathsf{d}(a))=\mathsf{d}_{1}(a)+\mathsf{d}_{2}(a)$

  4. (iv)

The adjoint zero map
$z^{\circ}_{A}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$, defined on generators as:

$\displaystyle z^{\circ}_{A}(a)=a$ $\displaystyle
z^{\circ}_{A}(\mathsf{d}(a))=0$

  5. (v)

The adjoint vertical lift
$\ell^{\circ}_{A}:{\mathsf{T}^{\circ}}^{2}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A)$, defined on
generators as:

$\displaystyle l^{\circ}_{A}(a)=a,$ $\displaystyle
l^{\circ}_{A}(\mathsf{d}(a))=0,$ $\displaystyle
l^{\circ}_{A}(\mathsf{d}^{\prime}(a))=0,$ $\displaystyle
l^{\circ}_{A}(\mathsf{d}^{\prime}\mathsf{d}(a))=\mathsf{d}(a).$

  6. (vi)

The adjoint canonical flip
$c^{\circ}_{A}:{\mathsf{T}^{\circ}}^{2}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathsf{T}^{\circ}}^{2}(A)$, defined on
generators as:

$\displaystyle c^{\circ}_{A}(a)=a,$ $\displaystyle
c^{\circ}_{A}(\mathsf{d}(a))=\mathsf{d}^{\prime}(a),$ $\displaystyle
c^{\circ}_{A}(\mathsf{d}^{\prime}(a))=\mathsf{d}(a),$ $\displaystyle
c^{\circ}_{A}(\mathsf{d}^{\prime}\mathsf{d}(a))=\mathsf{d}^{\prime}\mathsf{d}(a).$

  7. (vii)

The adjoint negative map
$n^{\circ}_{A}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}^{\circ}(A)$, defined on
generators as:

$\displaystyle n^{\circ}_{A}(a)=a$ $\displaystyle
n^{\circ}(\mathsf{d}(a))=-\mathsf{d}(a)$

Then,
$\mathbb{T}^{\circ}=(\mathsf{T}^{\circ},p^{\circ},s^{\circ},z^{\circ},l^{\circ},c^{\circ},n^{\circ})$
is a Rosický tangent structure on $\mathsf{ALG}^{op}_{\mathcal{P}}$, and so,
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ is a Cartesian Rosický
tangent category.

While the tangent bundle $\mathsf{T}$ is mostly the same for each operad, the
adjoint tangent bundle $\mathsf{T}^{\circ}$ can vary quite drastically from
operad to operad. So let us now consider the resulting tangent categories for
our main examples of operads. We again stress that, while the first two
examples recapture known examples, the last two examples are new examples of
tangent categories. While the third example is not too surprising, it does
provide a direct link between tangent categories and non-commutative algebraic
geometry, which provides a novel application for the theory of tangent
categories. On the other hand, the fourth example is a good example that
demonstrates how operads provide many new (and surprising) examples of tangent
categories that were not previously considered.

######  Example 4.4.5.

Let $A$ be an $R$-algebra. For the operad $A^{\bullet}$, Theorem 4.4.4
recaptures precisely the adjoint biproduct tangent structure from Lemma 3.3.1.
For an $A$-module $M$, every other $A$-module is an $M$-module in the operadic
sense, where the evaluation maps are all zero. Similarly, algebras over $M$ in
the operadic sense simply correspond to an $A$-module $N$ equipped with a
chosen $A$-linear morphisms $N\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M$. In this case,
$\mathsf{Free}_{M}(N)=M\times N$, which is an algebra over $M$ via the
injection map. On the other hand, $\Omega_{M}=M$, with universal derivation
being the identity $1_{M}:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M$. So, we indeed have that
$\mathsf{T}^{\circ}(M)=M\times M$. Thus
$(\mathsf{MOD}^{op}_{A},\mathbb{B}^{\circ})$ is a Cartesian Rosický tangent
category.

######  Example 4.4.6.

Recall that famously the opposite category of commutative $R$-algebras is
isomorphic to the category of affine schemes over $R$. Therefore, the
resulting tangent category for operad ${\operatorname{Com}}$ is equivalent to
the tangent category of affine schemes as described in [12, Section 2.3],
providing a link between tangent categories and algebraic geometry. For a
commutative $R$-algebra $A$, a module over $A$ in the operadic sense
corresponds precisely to a (left) $A$-module $M$. Free $A$-algebras are
constructed by the symmetric $A$-algebra functor:

$\mathsf{Free}_{A}(M)=\mathsf{Sym}_{A}(M)=\bigoplus_{n\in\mathbb{N}}\left(M^{\otimes_{A}n}\right)_{\Sigma(n)},$

where $\otimes_{A}$ is the tensor product over $A$. On the other hand,
$\Omega_{A}$ is precisely the usual (left) module of Kähler differentials over
$A$, that is, the free $A$-module over the set $\\{\mathsf{d}(a)|~{}\forall
a\in A\\}$ modulo the necessary derivation identities. Then
$\mathsf{T}^{\circ}(A)$ is the free symmetric $A$-algebra over $\Omega_{A}$:

$\mathsf{T}^{\circ}(A):=\mathsf{Sym}_{A}\left(\Omega_{A}\right)=\bigoplus\limits_{n=0}^{\infty}\left(\Omega_{A}^{\otimes_{A}n}\right)_{\Sigma(n)}=A\oplus\Omega_{A}\oplus\left(\Omega_{A}\otimes_{A}\Omega_{A}\right)_{\Sigma(2)}\oplus\ldots$

In [19, Definition 16.5.12.I], Grothendieck calls $\mathsf{T}^{\circ}(A)$ the
“fibré tangent” (french for tangent bundle) of $A$, while in [24, Section
2.6], Jubin calls $\mathsf{T}^{\circ}(A)$ the tangent algebra of $A$. An
arbitrary element of $\mathsf{T}^{\circ}(A)$ is a finite sum of monomials of
the form $a\mathsf{d}(b_{1})\ldots\mathsf{d}(b_{n})$, and thus the $R$-algebra
structure of $\mathsf{T}^{\circ}(A)$ is essentially the same as polynomials.
In particular, this implies that as an $R$-algebra, $\mathsf{T}^{\circ}(A)$ is
generated by $a$ and $\mathsf{d}(a)$ for all $a\in A$. On the other hand,
$\mathsf{T}_{n}(A)$ can be described in terms of generators $a$ and
$\mathsf{d}_{i}(a)$ for all $a\in A$ and $1\leq i\leq n$, and
${\mathsf{T}^{\circ}}^{2}(A)$ in terms of four generators $a$,
$\mathsf{d}(a)$, $\mathsf{d}^{\prime}(a)$, and
$\mathsf{d}^{\prime}\mathsf{d}(a)$ for all $a\in A$, modulo all the necessary
equations. The rest of the adjoint tangent structure is given as follows on
generators:

$\displaystyle p^{\circ}_{A}(a)$ $\displaystyle=a,$ $\displaystyle
q^{\circ}_{j}(a)$ $\displaystyle=a,$ $\displaystyle
q_{j}^{\circ}(\mathsf{d}(a))$ $\displaystyle=\mathsf{d}_{j}(a),$
$\displaystyle s^{\circ}_{A}(a)$ $\displaystyle=a,$ $\displaystyle
s^{\circ}(\mathsf{d}(a))$ $\displaystyle=\mathsf{d}_{1}(a)+\mathsf{d}_{2}(a),$
$\displaystyle z^{\circ}(a)$ $\displaystyle=a,$ $\displaystyle
z^{\circ}(\mathsf{d}(a))$ $\displaystyle=0,$ $\displaystyle l^{\circ}_{A}(a)$
$\displaystyle=a,$ $\displaystyle l^{\circ}_{A}(\mathsf{d}(a))$
$\displaystyle=0,$ $\displaystyle l^{\circ}_{A}(\mathsf{d}^{\prime}(a))$
$\displaystyle=0,$ $\displaystyle
l^{\circ}_{A}(\mathsf{d}^{\prime}\mathsf{d}(a))$
$\displaystyle=\mathsf{d}(a),$ $\displaystyle c^{\circ}_{A}(a)$
$\displaystyle=a,$ $\displaystyle c^{\circ}_{A}(\mathsf{d}(a))$
$\displaystyle=\mathsf{d}^{\prime}(a),$ $\displaystyle
c^{\circ}_{A}(\mathsf{d}^{\prime}(a))$ $\displaystyle=\mathsf{d}(a),$
$\displaystyle c^{\circ}_{A}(\mathsf{d}^{\prime}\mathsf{d}(a))$
$\displaystyle=\mathsf{d}^{\prime}\mathsf{d}(a),$ $\displaystyle
n^{\circ}_{A}(a)$ $\displaystyle=a,$ $\displaystyle n^{\circ}(\mathsf{d}(a))$
$\displaystyle=-\mathsf{d}(a).$

So $(\mathsf{CALG}^{op}_{R},\mathbb{T}^{\circ})$ is a Cartesian Rosický
tangent category.

######  Example 4.4.7.

For the operad ${\operatorname{Ass}}$, this results in a non-commutative
version of the previous example. For an $R$-algebra $A$, a module over $A$ in
the operadic sense corresponds precisely to an $A$-bimodule $M$. Free
$A$-algebras are given by the $A$-tensor algebra:

$\mathsf{Free}_{A}(M)=\mathsf{Ten}_{A}(M)=\bigoplus_{n\in\mathbb{N}}M^{\otimes_{A}n}.$

The $A$-module $\Omega_{A}$ is the non-commutative version of the module of
Kähler differentials over $A$ [18, Section 10] (which, it is important to
note, is different from the commutative version). Therefore,

$\mathsf{T}^{\circ}(A)=\bigoplus\limits_{n\in\mathbb{N}}\Omega_{A}^{\otimes_{A}n}=A\oplus\Omega_{A}\oplus(\Omega_{A}\otimes_{A}\Omega_{A})\oplus\ldots$

More concretely, $\mathsf{T}^{\circ}(A)$ can be described as the free
$A$-algebra over the set $\\{\mathsf{d}(a)|~{}\forall a\in A\\}$ modulo
$\mathsf{d}(ab)=a\mathsf{d}(b)+\mathsf{d}(a)b$ and
$\mathsf{d}(ra+sb)=r\mathsf{d}(a)+s\mathsf{d}(b)$. So
$(\mathsf{ALG}^{op}_{R},\mathbb{T}^{\circ})$ is a Cartesian Rosický tangent
category. In [18, Definition 10.2.3], Ginzburg calls $\mathsf{T}^{\circ}(A)$
the “space of noncommutative differential forms of $A$”. To the best of our
knowledge, this is the first mention of a tangent category that relates
directly to non-commutative algebraic geometry.

######  Example 4.4.8.

For the operad ${\operatorname{Lie}}$, we obtain a new example of a tangent
category for Lie algebras. For a Lie algebra ${\mathfrak{g}}$, modules in the
operadic sense correspond to representations of ${\mathfrak{g}}$, which we
call ${\mathfrak{g}}$-representations and simply denote by their underlying
$R$-module $V$. Algebras over ${\mathfrak{g}}$ in the operadic sense
correspond to Lie algebras ${\mathfrak{g}}^{\prime}$ equipped with a Lie
algebra morphism ${\mathfrak{g}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathfrak{g}}^{\prime}$. So,
$\mathsf{Free}_{\mathfrak{g}}(V)$ is the free Lie algebra over
${\mathfrak{g}}$ of a ${\mathfrak{g}}$-representation $V$. On the other hand,
$\Omega_{\mathfrak{g}}$ is the free representation of ${\mathfrak{g}}$ over
the set $\mathsf{d}({\mathfrak{g}})=\\{\mathsf{d}(x)|~{}\forall
x\in{\mathfrak{g}}\\}$ modulo the relations
$\mathsf{d}(rx+sy)=s\mathsf{d}(x)+r\mathsf{d}(y)$ and
$\mathsf{d}\left([x,y]\right)=[\mathsf{d}(x),y]+[x,\mathsf{d}(y)]$ for all
$r,s\in R$ and $x,y\in{\mathfrak{g}}$. Hence,
$\mathsf{T}^{\circ}({\mathfrak{g}})$ can be concretely defined as the free Lie
algebra over the underlying set of ${\mathfrak{g}}$ and the set
$\mathsf{d}({\mathfrak{g}})$ modulo the same equalities as for $\Omega_{A}$,
and such that $[x,y]\in\mathsf{T}^{\circ}({\mathfrak{g}})$ is identified to
$[x,y]\in{\mathfrak{g}}$, which makes $\mathsf{T}^{\circ}({\mathfrak{g}})$ a
Lie algebra over ${\mathfrak{g}}$. Therefore
$(\mathsf{LIE}^{op}_{R},\mathbb{T}^{\circ})$ is a Cartesian Rosický tangent
category, which we stress is a new important example of a tangent category.

We conclude this section by mentioning that the construction of Theorem 4.4.4
is functorial. Indeed, every operad morphism
${f:\mathcal{P}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}}$ induces a functor
$\mathsf{ALG}^{op}_{f}:\mathsf{ALG}^{op}_{\mathcal{P}^{\prime}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{ALG}^{op}_{\mathcal{P}}$ which
is defined on objects and maps as $\mathsf{ALG}_{f}$. It is straightforward to
check that
$\mathsf{ALG}^{op}_{f}:(\mathsf{ALG}^{op}_{\mathcal{P}^{\prime}},\mathbb{T}^{\circ})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$
is a Cartesian Rosický tangent morphism. Now let $\mathsf{CRTAN}$ be the
category of Cartesian Rosický tangent categories and Cartesian tangent
morphisms between them [6, Section 4.3]. Therefore we obtain a functor
$\mathsf{ALG}^{op}:\mathsf{OPERAD}^{op}_{R}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{CRTAN}$ which sends an operad
$\mathcal{P}$ to $(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ and an
operad morphism $f:\mathcal{P}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}^{\prime}$ to
$\mathsf{ALG}^{op}_{f}:(\mathsf{ALG}^{op}_{\mathcal{P}^{\prime}},\mathbb{T}^{\circ})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$.
We note that the Cartesian tangent morphism $\mathsf{ALG}^{op}_{f}$ is not
strong.

###  4.5 Vector Fields of Algebras of an Operad

In this section, we will explain how in the category of algebras of an operad,
vector fields correspond precisely to derivations. Luckily, it turns out that
it is already known that derivations are closely related to the semi-direct
product, i.e., the tangent bundle. Indeed, one could apply [28, Proposition
12.3.11] to get the desired result. However, let us give an alternative
explanation using the coCartesian differential monad point of view.

So let $\mathcal{P}$ be an operad and let $A$ be a $\mathcal{P}$-algebra. In
Section 3.4 we explained how vector field of $A$ in
$(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ correspond to
$\mathsf{S}(\mathcal{P},-)$-derivations of $A$. It turns out that
$\mathsf{S}(\mathcal{P},-)$-derivations on $A$ correspond precisely to
$A$-derivations evaluated in itself. Indeed, $A$ is an $A$-module where the
evaluation maps are induced from the $\mathcal{P}$-algebra structure
$\theta:\mathsf{S}(\mathcal{P},A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$. Then we may consider $A$-derivations
$D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$.

######  Lemma 4.5.1.

For an operad $\mathcal{P}$ and a $\mathcal{P}$-algebra $A$, an
$\mathsf{S}(\mathcal{P},-)$-derivation $D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ is precisely the same as a
$A$-derivation $D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$.

###### Proof.

Recall that an $R$-linear morphism $D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ is an $A$-derivation if it satisfies:

$D(\mu(a_{1},\dots,a_{n}))=\sum_{i=1}^{n}\mu(a_{1},\ldots,D(a_{i}),\ldots,a_{n}).$

On the other hand, an $R$-linear morphism
$D:A\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}A$
is an $\mathsf{S}(\mathcal{P},-)$-derivation if $D\circ\theta=\theta\circ
S\left(\mathcal{P},\pi_{1}+D\circ\pi_{2}\right)\circ\partial_{A}$. Let us show
that this equality is precisely the same as requiring $D$ be an
$A$-derivation. For $\mu_{A}\in\mathcal{P}(0)$, since
$\partial_{A}(\mu_{A})=0$, we have that $D(\mu_{A})=0$. For the rest, we
compute:

$\displaystyle D(\mu(a_{1},\dots,a_{n}))$
$\displaystyle=D\left(\theta(\mu;a_{1},\ldots,a_{n})\right)=\theta\left(S\left(\mathcal{P},\pi_{1}+D\circ\pi_{2}\right)\left(\partial_{A}(\mu;a_{1},\dots,a_{n})\right)\right)$
$\displaystyle=\sum_{i=1}^{n}\theta\left(S\left(\mathcal{P},\pi_{1}+D\circ\pi_{2}\right)\left(\mu;(a_{1},0),\ldots,(0,a_{i}),\ldots,(a_{n},0)\right)\right)$
$\displaystyle=\sum_{i=1}^{n}\theta\left(\mu;a_{1},\ldots,D(a_{i}),\ldots,a_{n}\right)=\sum_{i=1}^{n}\mu(a_{1},\ldots,D(a_{i}),\ldots,a_{n}).$

We conclude that $\mathsf{S}(\mathcal{P},-)$-derivations on $A$ and
$A$-derivations evaluated in $A$ are indeed the same thing. ∎

Therefore, by Lemma 3.4.2, we conclude that vector fields correspond to
derivations as desired:

######  Proposition 4.5.2.

For an operad $\mathcal{P}$ and a $\mathcal{P}$-algebra $A$, there is a
bijective correspondence between vector fields of $A$ in
$(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ and $A$-derivations
$D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$. Therefore a vector field
$v\in\mathsf{V}_{\mathbb{T}}(A)$ is precisely a $\mathcal{P}$-algebra morphism
$v:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A\ltimes A$ such that
$v(a)=(a,D_{v}(a))$ for all $a\in A$, where
$D_{v}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ is an $A$-derivation. Furthermore,
the induced Lie bracket is given by
$[v,w](a)=(a,D_{v}(D_{w}(a))-D_{w}(D_{v}(a)))$.

By Lemma 2.3.3, we also have that vector fields in the opposite category of
algebras also correspond precisely to derivations:

######  Corollary 4.5.3.

For an operad $\mathcal{P}$ and a $\mathcal{P}$-algebra $A$, there is a
bijective correspondence between vector fields of $A$ in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ and $A$-derivations
$D:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$. So a vector field
$v\in\mathsf{V}_{\mathbb{T}^{\circ}}(A)$ is precisely a $\mathcal{P}$-algebra
morphism $v:\mathsf{Free}_{A}(\Omega_{A})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ which is defined on generators as
$v(a)=a$ and $v(\mathsf{d}(a))=D_{v}(a)$ for all $a\in A$, where
${D_{v}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A}$ is an $A$-derivation. Furthermore,
the induced Lie bracket is given on generators by $[v,w](a)=a$ and
$[v,w](\mathsf{d}(a))=D_{v}(D_{w}(a))-D_{w}(D_{v}(a))$.

Let us consider what vector fields are for our main examples of operads.

######  Example 4.5.4.

For an $R$-algebra $A$ and an $A$-module $M$, an $M$-derivation evaluated in
$M$ is just an $A$-linear endomorphism $f:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M$. Therefore a vector field of $M$ in
$(\mathsf{MOD}_{A},\mathbb{B})$ is an $A$-linear map
$v:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M\times M$ such that $v(m)=(m,f_{v}(m))$
for some $A$-linear map $f_{v}:M\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}M$. Similarly for vector fields of $M$
in $(\mathsf{MOD}^{op}_{A},\mathbb{B}^{\circ})$.

######  Example 4.5.5.

For the operad ${\operatorname{Com}}$ and a commutative $R$-algebra $A$, an
$A$-derivation evaluated in $A$ in the operadic sense is the same thing as a
derivation in the classical sense, that is, $R$-linear morphism
$D:A\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}A$
which satisfies the product rule: $D(ab)=aD(b)+D(a)b$. Then a vector field of
$A$ in $(\mathsf{CALG}_{R},\mathbb{T})$ is an $R$-algebra morphism
$v:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A[\epsilon]$ such that
$v(a)=a+D_{v}(a)\epsilon$ for some derivation
$D_{v}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$. Similarly, a vector field of $A$ in
$(\mathsf{CALG}^{op}_{R},\mathbb{T}^{\circ})$ corresponds to an $R$-algebra
morphism
$v:\mathsf{Sym}_{A}\left(\Omega_{A}\right)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ which is given on generators as
$v(a)=a$ and $v(\mathsf{d}(a))=D_{v}(a)$ for some derivation
$D_{v}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$.

######  Example 4.5.6.

For the operad ${\operatorname{Ass}}$, derivations in the operadic sense again
correspond to derivations in the classical sense as in the previous example.
So vector fields in $(\mathsf{ALG}_{R},\mathbb{T})$ or
$(\mathsf{ALG}^{op}_{R},\mathbb{T}^{\circ})$ are given in essentially the same
way as in the commutative case.

######  Example 4.5.7.

For the operad ${\operatorname{Lie}}$ and a Lie algebra ${\mathfrak{g}}$, a
${\mathfrak{g}}$-derivation evaluated in ${\mathfrak{g}}$ corresponds to an
$R$-linear mormphsim $D:{\mathfrak{g}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathfrak{g}}$ which satisfies
$D([x,y])=[x,D(y)]+[D(x),y]$ for all $x,y\in{\mathfrak{g}}$. So a vector field
of ${\mathfrak{g}}$ in $(\mathsf{LIE}_{R},\mathbb{T})$ is a Lie algebra
morphism $v:{\mathfrak{g}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathfrak{g}}\ltimes{\mathfrak{g}}$
such that $v(x)=x+D_{v}(y)\epsilon$ for some ${\mathfrak{g}}$-derivation
$D_{v}:{\mathfrak{g}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathfrak{g}}$. Similarly, a vector
field of ${\mathfrak{g}}$ in $(\mathsf{Lie}^{op}_{R},\mathbb{T}^{\circ})$
corresponds to an $R$-algebra morphism
$v:\mathsf{T}^{\circ}({\mathfrak{g}})\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ which is given on generators as
$v(a)=a$ and $v(\mathsf{d}(a))=D_{v}(a)$ for some ${\mathfrak{g}}$-derivation
$D_{v}:{\mathfrak{g}}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}{\mathfrak{g}}$.

###  4.6 Differential Objects of an Operad

In this section, we will give precise characterizations of the differential
objects in both the category of algebras of an operad and its opposite
category. On the one hand, for the category of algebras, we will see that the
differential objects are in some sense quite trivial algebras. On the other
hand, we will show that the differential objects in the opposite category are
quite rich and recapture a certain kind of module in the operadic sense.

Let us begin by taking a look at the differential objects in the category of
algebras of an operad.

######  Proposition 4.6.1.

Let $\mathcal{P}$ be an operad. Then a $\mathcal{P}$-algebra $A$ is a
differential object in $(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ (in a
necessarily unique way) if and only if $\mu(a_{1},\ldots,a_{n})=0$ for all
$n\neq 1$, $\mu\in\mathcal{P}(n)$, and $a_{i}\in A$.

###### Proof.

Recall that the terminal object in $\mathsf{ALG}_{\mathcal{P}}$ is given by
the zero $R$-module $\mathsf{0}$, whose $\mathcal{P}$-algebra structure is
just given by $0$. Per the discussions in Section 3.5, a $\mathcal{P}$-algebra
$A$ is a differential object if and only if $\pi_{2}:A\ltimes
A\allowbreak\mathrel{\mathop{\hbox to11.99998pt{\rightarrowfill}}\limits}A$,
$\pi_{1}+\pi_{2}:A\times A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$, and
$0:\mathsf{0}\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ are all $\mathcal{P}$-algebra
morphisms. Suppose that $A$ is a differential object. Then, as mentioned in
Section 3.5, we have that $A\times A=A\ltimes A$. This means that:

$\displaystyle(\mu(a_{1},\ldots,a_{n}),\mu(b_{1},\ldots,b_{n}))$
$\displaystyle=\mu((a_{1},b_{1}),\ldots,(a_{n},b_{n}))$
$\displaystyle=\left(\mu(a_{1},\ldots,a_{n}),\sum_{i=1}^{n}\mu(a_{1},\ldots,b_{i},\ldots,a_{n})\right),$

which implies:

$\displaystyle\mu(b_{1},\ldots,b_{n})=\sum_{i=1}^{n}\mu(a_{1},\ldots,b_{i},\ldots,a_{n}),$
$\displaystyle\forall a_{i},b_{i}\in A.$

The left-hand side does not depend on $a_{i}$, so we get:
$\mu(b_{1},\ldots,b_{n})=\sum_{i=1}^{n}\mu(0,\ldots,b_{i},\ldots,0)$. But by
multilinearity, $\sum_{i=1}^{n}\mu(0,\ldots,b_{i},\ldots,0)=0$ for $n\geq 2$.
For $n=0$, we also get an empty sum. Therefore, $\mu(b_{1},\ldots,b_{n})=0$
for all $n\neq 1$. Conversely, if in $A$ $\mu(b_{1},\ldots,b_{n})=0$ for all
$n\neq 1$, it is straightforward to show that the equalities in Lemma 3.5.1
hold, and so, $A$ is a differential object. ∎

Here are the differential objects for our main examples of operads:

######  Example 4.6.2.

For any $R$-algebra $A$, $A^{\bullet}(1)=A$; every $A$-module $M$ is a
differential object in $(\mathsf{MOD}_{A},\mathbb{B})$, as per the discussion
in Section 3.5.

######  Example 4.6.3.

For the operad ${\operatorname{Com}}$, a commutative $R$-algebra $A$ is a
differential object in $(\mathsf{CALG}_{R},\mathbb{T})$ would imply that
$A[\epsilon]\cong A\times A$ as $R$-algebras. However, the unit in
$A[\epsilon]$ is $1$ while the unit in $A\times A$ is $(1,1)$. But then the
isomorphism $A[\epsilon]\cong A\times A$ would imply that $1=0$. This is only
the case for the zero $R$-algebra $\mathsf{0}$. So the only differential
object in $(\mathsf{CALG}_{R},\mathbb{T})$ is $\mathsf{0}$.

######  Example 4.6.4.

For the operad ${\operatorname{Ass}}$, by the same argument as in the previous
example, we have that the differential object in
$(\mathsf{ALG}_{R},\mathbb{T})$ is $\mathsf{0}$.

######  Example 4.6.5.

For the operad ${\operatorname{Lie}}$, it turns out that the differential
objects are precisely the $R$-modules. Indeed, every $R$-module $V$ comes
equipped with a trivial Lie bracket, $[v,w]=0$, which makes $V$ a Lie algebra
and also that $V\ltimes V=V\times V$. Conversely, suppose that
${\mathfrak{g}}$ is a Lie algebra and a differential object, which in
particular implies that
${\mathfrak{g}}\ltimes{\mathfrak{g}}={\mathfrak{g}}\times{\mathfrak{g}}$.
However, this implies that
$\left([x_{1},x_{2}],[y_{1},y_{2}]\right)=[(x_{1},y_{1}),(x_{2},y_{2})]=\left([x_{1},x_{2}],[x_{1},y_{2}]+[y_{1},x_{2}]\right)$.
Setting $x_{i}=0$, we get that $[y_{1},y_{2}]=0$, which means that the Lie
bracket of ${\mathfrak{g}}$ is trivial. So we have that the differential
objects in $(\mathsf{LIE}_{R},\mathbb{T})$ are precisely the $R$-modules with
the trivial Lie bracket.

Let us now turn our attention to differential objects in the opposite category
of algebras of an operad $\mathcal{P}$. Luckily, as mentioned in Section 2.4,
differential objects do not transfer through adjoint tangent structure. So
even if $(\mathsf{ALG}_{\mathcal{P}},\mathbb{T})$ may not have any non-trivial
differential objects, we will show that
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ actually has many
interesting differential objects. Recall that we mentioned that
$\mathsf{ALG}_{\mathcal{P}}$ is cocomplete [30, Proposition 6.4], and
therefore has coproducts. However, coproducts of $\mathcal{P}$-algebras are
not straightforward and easy to work with. Luckily, there is an alternative
but equivalent characterization of differential objects in a Cartesian Rosický
tangent category which does not involve the product $\times$. As such, this
alternative description will allow us to describe differential objects in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ without having to work
with the coproduct in $\mathsf{ALG}_{\mathcal{P}}$.

Firstly, it turns out that a differential object is in fact a special kind of
differential bundle [9, Definition 2.3], which are analogues of smooth vector
bundles in a tangent category. While differential bundles are beyond the scope
of this paper (we invite interested readers to learn about them in [9, 12,
29]), it is enough to know that a differential object is the same thing as a
differential bundle over the terminal object $\ast$ [9, Proposition 3.4]. In
[29], MacAdam provided an alternative description of a differential bundle in
a Cartesian Rosický tangent category, which in particular required less data
and axioms than the original definition. Briefly, MacAdam showed that, in a
Cartesian Rosický tangent category, a differential bundle over an object $X$
can be characterized as an object $A$ with maps
$q:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}X$, called the projection,
$\zeta:X\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$, called the zero map, and
$\ell:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}(A)$, called the lift map.
Furthermore, these need to satisfy: (1) four equalities, (2) that the pullback
of $n$-copies of $q$ exists and is preserved by $\mathsf{T}^{n}$, and (3)
$\ell$ has a universal property being part of a pullback square called the
Rosický’s universality diagram, see [29, Corollary 3] or [12, Proposition 3.8]
for full details. We have mentioned that differential objects are just
differential bundles over the terminal object. So, in a Cartesian Rosický
tangent category, MacAdam’s description is greatly simplified when $X=\ast$.
Firstly, there is only one possible candidate for the projection: the unique
map $t_{A}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\ast$, so this map need not be
specified. It follows that condition (2) is just saying that the product of
$n$ copies of $A$ exists and is preserved by the tangent bundle, which is
already true since we are in a Cartesian tangent category. So, condition (2)
is automatically verified. So, one only needs the relations on the maps
$\zeta:\ast\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ and
${\ell:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}(A)}$. Now, one of the
equalities in condition (1) is $t_{A}\circ\zeta=1_{\ast}$, which is true by
the universal property of the terminal object, so it is always satisfied.
Lastly, the pullback square of condition (3) usually has
$\mathsf{T}(\ast)\times A$ in the bottom corner. However, since
$\mathsf{T}(\ast)\cong\ast$, we may rewrite Rosický’s universality diagram
with $A$ in the bottom corner. Therefore, MacAdam’s description allows us to
provide a much simpler, yet equivalent, characterization of differential
objects without referring to products.

The following is just a combination of [9, Proposition 3.4] with the rewritten
version of [29, Corollary 3], or [12, Proposition 3.8], for the specific case
of the terminal object.

######  Lemma 4.6.6.

Let $(\mathbb{X},\mathbb{T})$ be a Cartesian Rosický tangent category. Then,
there is a bijective correspondence between:

  1. (i)

Differential objects $(A,\hat{p},\sigma,\zeta)$;

  2. (ii)

Triples $(A,\zeta,\ell)$ consisting of an object $A$, a map
$\zeta:\ast\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$, called the zero map, and a map
$\ell:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{T}(A)$, called the differential
lift, such that the following equalities hold:

$\displaystyle p_{A}\circ\ell=\zeta\circ t_{A},$
$\displaystyle\ell\circ\zeta=z_{A}\circ\zeta,$
$\displaystyle\mathsf{T}(\ell)\circ\ell=l_{A}\circ\ell,$ (2)

and the following commutative diagram, called Rosický’s universality diagram,
is a pullback square:

$\begin{gathered}\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
11.52779pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
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0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\ell}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 66.75pt\raise 0.0pt\hbox{\hbox{\kern
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0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
66.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
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3.0pt}}}}}}\ignorespaces{\hbox{\kern
81.00002pt\raise-28.58334pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\ast\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
35.34377pt\raise-44.52776pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\zeta}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
74.25002pt\raise-38.41666pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
74.25002pt\raise-38.41666pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A}$}}}}}}}\ignorespaces}}}}\ignorespaces\end{gathered}$

and for all $n\in\mathbb{N}$, $\mathsf{T}^{n}$ preserves this pullback.

In order to use Lemma 4.6.6, we must mention what the terminal object is in
$\mathsf{ALG}^{op}_{\mathcal{P}}$. Firstly note that $\mathcal{P}(0)$ is a
$\mathcal{P}$-algebra where the $\mathcal{P}$-algebra structure is just given
by the composition operation of $\mathcal{P}$. Furthermore, $\mathcal{P}(0)$
is the initial object in $\mathsf{ALG}_{\mathcal{P}}$, that is, for every
$\mathcal{P}$-algebra $A$ there is a unique $\mathcal{P}$-algebra morphism
$t^{\circ}_{A}:\mathcal{P}(0)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$, which is defined as
$t_{A}(\mu)=\mu_{A}$. Thus $\mathcal{P}(0)$ is the terminal object in
$\mathsf{ALG}^{op}_{\mathcal{P}}$.

So by Lemma 4.6.6, a differential object in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ is a
$\mathcal{P}$-algebra $A$ equipped with $\mathcal{P}$-algebra morphisms
$\zeta^{\circ}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}(0)$ and
$\ell^{\circ}:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ such that the dual of equalities in
(2) hold and the following square is a pushout diagram in
$\mathsf{ALG}_{\mathcal{P}}$:

$\begin{gathered}\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
12.79169pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
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33.5099pt\raise 5.19028pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.82361pt\hbox{$\scriptstyle{p_{A}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 72.79169pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.19444pt\hbox{$\scriptstyle{\zeta^{\circ}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 0.0pt\raise-29.78888pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
72.79169pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\mathsf{T}^{\circ}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
88.44171pt\raise-17.97778pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.875pt\hbox{$\scriptstyle{\ell^{\circ}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
88.44171pt\raise-30.45557pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx<EMAIL_ADDRESS>0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{\mathcal{P}(0)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
41.97781pt\raise-46.56943pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
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3.0pt}}}}}}\ignorespaces{\hbox{\kern
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0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
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0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A}$}}}}}}}\ignorespaces}}}}\ignorespaces\end{gathered}$

Note that we do not need to mention that ${\mathsf{T}^{\circ}}^{n}$ preserves
these pushout squares, since $\mathsf{T}^{\circ}$ is a left adjoint, and left
adjoints always preserve colimits. Let us unpack this a bit more. For starters
for every $a\in A$, $\zeta^{\circ}(a)\in\mathcal{P}(0)$ so we have
$\zeta^{\circ}(a)_{A}\in A$. Therefore the dual of equalities in (2) amount to
the following identities on generators:

$\begin{gathered}\ell^{\circ}(a)=\zeta^{\circ}(a)_{A},\\\
\zeta^{\circ}(\zeta^{\circ}(a)_{A})_{A}=\zeta^{\circ}(a)_{A}\qquad\zeta^{\circ}(\ell^{\circ}(\mathsf{d}(a)))_{A}=0,\\\
\ell^{\circ}(\zeta^{\circ}(a)_{A})=\zeta^{\circ}(a)_{A}\qquad\ell^{\circ}(\mathsf{d}\left(\ell^{\circ}(\mathsf{d}(a))\right))=\ell^{\circ}(\mathsf{d}(a)).\end{gathered}$
(3)

Lastly, the pushout property says that for any $\mathcal{P}$-algebra
$A^{\prime}$ and $\mathcal{P}$-algebra morphism
$f:\mathsf{T}^{\circ}(A)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$ such that
$f(a)=\zeta^{\circ}(a)_{A}$, there exists a unique $\mathcal{P}$-algebra
morphism $f^{\natural}:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$ such that
$f^{\natural}(\ell(x))=f(x)$ for all $x\in\mathsf{T}^{\circ}(A)$, so in
particular on generators
$f^{\natural}(\zeta^{\circ}(a)_{A})=\zeta^{\circ}(a)_{A^{\prime}}$ and
$f^{\natural}(\ell^{\circ}(\mathsf{d}(a)))=f(\mathsf{d}(a))$.

By Proposition 3.5.2, we already know that every free $\mathcal{P}$-algebra is
a differential object:

######  Lemma 4.6.7.

Let $\mathcal{P}$ be an operad. Then for any $R$-module $V$,
$\mathsf{S}(\mathcal{P},V)$ is a differential object in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ where in particular the
zero $\zeta^{\circ}:\mathsf{S}(\mathcal{P},V)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}(0)$ is defined as follows:

$\displaystyle\zeta^{\circ}(\mu;v_{1},\ldots,v_{n})=0,$

and the differential lift
$\ell^{\circ}:\mathsf{T}^{\circ}(\mathsf{S}(\mathcal{P},V))\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{S}(\mathcal{P},V)$ is defined as
follows on generators:

$\displaystyle\ell^{\circ}(\mu;v_{1},\ldots,v_{n})$ $\displaystyle=0,$
$\displaystyle\ell^{\circ}\left(\mathsf{d}(\mu;v_{1},\ldots,v_{n})\right)$
$\displaystyle=(\mu;v_{1},\ldots,v_{n}).$

Furthermore,
$\mathsf{T}^{\circ}(\mathsf{S}(\mathcal{P},V))\cong\mathsf{S}(\mathcal{P},V\times
V)$ as $\mathcal{P}$-algebras.

While free $\mathcal{P}$-algebras always give differential objects, it is
possible that there are other differential objects. That said, we are still
able to precisely characterize the differential objects as
$\mathcal{P}(0)$-modules (in the operadic sense).

######  Theorem 4.6.8.

Let $\mathcal{P}$ be an operad. Then there is a bijective correspondence
between differential objects in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$ and
$\mathcal{P}(0)$-modules.

###### Proof.

Let $(A,\zeta^{\circ},\ell^{\circ})$ be a differential object in
$(\mathsf{ALG}^{op}_{\mathcal{P}},\mathbb{T}^{\circ})$. Let
$\mathsf{D}_{\ell^{\circ}}(a)=\ell^{\circ}(\mathsf{d}(a))$ and let
$\mathsf{D}_{\ell^{\circ}}(A)$ be the image of $\mathsf{D}_{\ell^{\circ}}$, so
$\mathsf{D}_{\ell^{\circ}}(A)$ is an $R$-module. We claim that the following
equips $\mathsf{D}_{\ell^{\circ}}(A)$ with a $\mathcal{P}(0)$-module structure
with evaluation:

$\displaystyle\psi_{n+1}\left(\mu;\nu_{1},\ldots,\nu_{n},\ell^{\circ}(\mathsf{d}(a)))\right):=\ell^{\circ}\left(\mathsf{d}\left(\mu({\nu_{1}},\ldots,{\nu_{n}},a)\right)\right).$

Indeed, by definition, since $\mathsf{d}$ is a derivation, one has:

$\displaystyle\ell^{\circ}\left(\mathsf{d}\left(\mu({\nu_{1}},\ldots,{\nu_{n}},a)\right)\right)$
$\displaystyle=\ell^{\circ}\left(\sum_{i=1}^{n}\mu({\nu_{1}},\ldots,\mathsf{d}\nu_{i},\ldots{\nu_{n}},a)\right)+\ell^{\circ}\left(\mu({\nu_{1}},\ldots,{\nu_{n}},\mathsf{d}a)\right)$
$\displaystyle=\mu\left({\nu_{1}},\ldots,{\nu_{n}},\ell^{\circ}(\mathsf{d}a)\right),$

which gives a $\mathcal{P}(0)$-module structure on
$\mathsf{D}_{\ell^{\circ}}(A)$.

Conversely, let $M$ be a $\mathcal{P}(0)$-module. Consider the free
$\mathcal{P}(0)$-algebra $\mathsf{Free}_{\mathcal{P}(0)}(M)$. As shown in
Proposition 4.4.1, $\mathsf{Free}_{\mathcal{P}(0)}(M)$ is generated by
$\mu\in\mathcal{P}(0)$ and $x\in M$. However, thanks to the relations on
$\mathsf{Free}_{\mathcal{P}(0)}(M)$, the generators $\mu\in\mathcal{P}(0)$
correspond in fact to the units coming from the $\mathcal{P}$-algebra
structure, so $\mathsf{Free}_{\mathcal{P}(0)}(M)$, as $\mathcal{P}(0)$-algebra
is generated by $x\in M$. The $\mathcal{P}$-algebra
$\mathsf{T}^{\circ}(\mathsf{Free}_{\mathcal{P}(0)}(M))$ has generators $x$,
and $\mathsf{d}(x)$ for all $x\in M$. Now, define the differential lift
$\ell^{\circ}:\mathsf{T}^{\circ}(\mathsf{Free}_{\mathcal{P}(0)}(M))\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{\mathcal{P}(0)}(M)$ as
the $\mathcal{P}$-algebra morphism defined as follows on generators:

$\displaystyle\ell^{\circ}(x)$ $\displaystyle=0,$
$\displaystyle\ell^{\circ}(\mathsf{d}(x))$ $\displaystyle=x,$

which is indeed well-defined, since $\ell^{\circ}$ can be constructed using
the universal properties. Next, we define the zero
${\zeta^{\circ}:\mathsf{Free}_{\mathcal{P}(0)}(M)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathcal{P}(0)}$ as the
$\mathcal{P}$-algebra morphism defined as follows on generators:

$\zeta^{\circ}(x)=0.$

It is straightforward to check that $\ell^{\circ}$ and $\zeta^{\circ}$ satisfy
the equalities of (3). Lastly for the pushout square, suppose that there was a
$\mathcal{P}$-algebra morphism
$f:\mathsf{T}^{\circ}(\mathsf{Free}_{\mathcal{P}(0)}(M))\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$ such that $f(x)=0$. Then
define the $\mathcal{P}$-algebra morphism
$f^{\natural}:\mathsf{Free}_{\mathcal{P}(0)}(M)\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A^{\prime}$ on generators as
$f^{\natural}(x)=f(\mathsf{d}(x))$. By construction, $f^{\natural}$ satisfies
the necessary identities and is the unique map that does so since it was
defined on generators. So we conclude that
$(\mathsf{Free}_{\mathcal{P}(0)}(M),\zeta^{\circ},\ell^{\circ})$ is indeed a
differential object.

We must now explain why these constructions are inverses of each other. So
starting from a $\mathcal{P}(0)$-module $M$, by definition of $\ell^{\circ}$
we already have that
$\mathsf{D}_{\ell^{\circ}}(\mathsf{Free}_{\mathcal{P}(0)}(M))=M$ (and this is
an equality of $\mathcal{P}(0)$-modules). In the other direction, start with a
differential object $(A,\zeta^{\circ},\ell^{\circ})$. Define the
$\mathcal{P}$-algebra morphism $\phi:A\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}\mathsf{Free}_{\mathcal{P}(0)}(\mathsf{D}_{\ell^{\circ}}(A))$
using the universal property of the pushout as the unique
$\mathcal{P}$-algebra morphism such that:

$\phi(\ell^{\circ}(\mathsf{d}(a)))=\ell^{\circ}(\mathsf{d}(a)).$

Then, $\phi$ is a $\mathcal{P}$-algebra isomorphism with inverse
$\phi^{-1}:\mathsf{Free}_{\mathcal{P}(0)}(\mathsf{D}_{\ell^{\circ}}(A))\allowbreak\mathrel{\mathop{\hbox
to11.99998pt{\rightarrowfill}}\limits}A$ defined on generators as:

$\phi^{-1}(\ell^{\circ}(\mathsf{d}(a)))=\ell^{\circ}(\mathsf{d}(a)).$

We have that
$A\cong\mathsf{Free}_{\mathcal{P}(0)}(\mathsf{D}_{\ell^{\circ}}(A))$ as
$\mathcal{P}$-algebras, and it easy to see that $\phi$ preserves the
differential object structure $\zeta^{\circ}$ and $\ell^{\circ}$. So, we
conclude that there is indeed a bijective correspondence between differential
objects and $\mathcal{P}(0)$-modules as desired. ∎

Before considering the differential objects in our main examples, we point out
that $\mathcal{P}(0)$-modules also have an alternative description in terms of
modules in the usual sense.

######  Lemma 4.6.9.

[2, Lemma 1.4] For an operad $\mathcal{P}$, $\mathcal{P}(0)$-modules in the
operadic sense correspond precisely to $\mathcal{P}(1)$-left modules in the
usual sense.

Let us consider the differential objects in our main examples of operads. In
particular, we note that the first example has differential objects which are
not simply free $\mathcal{P}$-algebras. On the other hand, for the last three
examples, the differential objects turn out to be precisely free
$\mathcal{P}$-algebras.

######  Example 4.6.10.

For an $R$-algebra $A$, $A^{\bullet}(1)=A$; so every $A$-module is a
differential object in $(\mathsf{MOD}^{op}_{A},\mathbb{B}^{\circ})$, as per
the discussion in Section 3.5.

######  Example 4.6.11.

For the operad ${\operatorname{Com}}$, ${\operatorname{Com}}(0)=R$ and
${\operatorname{Com}}(1)=R$. We also have that
$\mathsf{Free}_{R}=\mathsf{Sym}$. Therefore for any $R$-module $V$,
$\mathsf{Sym}(V)$ is a differential object in
$(\mathsf{CALG}^{op}_{R},\mathbb{T}^{\circ})$. If $V$ is a free $R$-module
with basis $X$, then recall that $\mathsf{Sym}(V)\cong R[X]$ and so
$\mathsf{T}^{\circ}(\mathsf{Sym}(V))\cong R[X,dX]$, as in Example 4.2.4. So in
terms of polynomials, the differential object structure is defined as follows

$\displaystyle\hat{p}^{\circ}(q(x_{1},\ldots,x_{n}))$
$\displaystyle=q(dx_{1},\ldots,dx_{n}),$
$\displaystyle\zeta^{\circ}(q(x_{1},\ldots,x_{n}))$
$\displaystyle=q(0,\ldots,0),$
$\displaystyle\sigma^{\circ}(q(x_{1},\ldots,x_{n}))$
$\displaystyle=q(x_{1}+dx_{1},\ldots,x_{n}+dx_{n}),$
$\displaystyle\ell^{\circ}(q(x_{1},\ldots,x_{n},dy_{1},\ldots,dx_{m}))$
$\displaystyle=q(0,\ldots,0,dy_{1},\ldots,dx_{m}).$

This recaptures [12, Theorem 5.9].

######  Example 4.6.12.

For the operad ${\operatorname{Ass}}$, ${\operatorname{Ass}}(0)=R$ and
${\operatorname{Ass}}(1)=R$. We also have that
$\mathsf{Free}_{R}=\mathsf{Ten}$. So for any $R$-module $V$, $\mathsf{Ten}(V)$
is a differential object in $(\mathsf{ALG}^{op}_{R},\mathbb{T}^{\circ})$. For
the case of free $R$-modules, we may describe the differential object
structure in terms of non-commutative polynomials as in the previous example.

######  Example 4.6.13.

For the operad ${\operatorname{Lie}}$, ${\operatorname{Lie}}(0)=\mathsf{0}$
and ${\operatorname{Lie}}(1)=R$. We also have
$\mathsf{Free}_{\mathsf{0}}=\mathsf{Lie}$. Thus for any $R$-module $V$, the
free Lie algebra $\mathsf{Lie}(V)$ is a differential object in
$(\mathsf{Lie}^{op}_{R},\mathbb{T}^{\circ})$.

##  5 Future Work

We conclude this paper by discussing some interesting future research projects
that build upon the theory of tangent categories of algebras over an operad.

  1. (i)

In [12], it was shown that the study of the tangent structure on the opposite
category of commutative algebras provides many concepts from the algebraic
geometry of affine schemes. Similarly, the tangent structure on the opposite
category of associative algebras formalizes many constructions of Ginzburg
[18] related to non-commutative geometry. It is natural to ask the question:
what kind of geometry can be described using the opposite category of Lie
algebras? Is this somehow related to the geometry of Lie algebras studied, for
example, by Francis and Gaitsgory [14]? What are the geometries obtained using
the operads of PreLie algebras or Poisson algebras? Can we describe the non-
commutative Poisson geometry studied by Van den Bergh [35] using our
techniques? Even more generally, one could use tangent categories to provide a
new version of algebraic geometry relative to an operad.

  2. (ii)

Differential bundles in a tangent category [9] generalize the notion of smooth
vector bundles. In [12], it was shown that differential bundles over a
commutative algebra correspond precisely to modules over said commutative
algebra. As such, we conjecture the differential bundles over an algebra of an
operad will also correspond to modules over the algebra (in the operadic
sense). This would be an extension of Theorem 2.2.3 since a differential
object can equivalently be described as a differential bundle over the
terminal object. This will be investigated by the second named author for
their PhD thesis. Beyond differential bundles, one should also study other
interesting tangent category notions in the (opposite) category of algebras
over an operad. For example, what are the tangent category versions of
connections [8], or de Rham cohomology [13], or even solving differential
equations [10] in these tangent categories?

  3. (iii)

The story of this paper was to explain how, from operad, one could build
tangent categories and Cartesian differential categories. A natural question
is if we can go in the other direction. So for what kinds of tangent
categories or Cartesian differential categories is possible to construct an
operad? Similarly, it would be of interest to precisely characterize which
tangent categories are equivalent to the (opposite) category of algebras of an
operad.

  4. (iv)

In [1], Bauer, Burke and Ching generalized the notion of tangent categories to
the higher categorical setting. This new concept of tangent $\infty$-category
allows one to study tangent structures “up to higher coherences”. There is a
well-established notion of $\infty$-operad [5], and this encodes operations
“up to homotopy”. It then seems natural to ask whether our theory can be
generalized to produce a tangent $\infty$-structure on the (opposite) category
of algebras over an $\infty$-operad. If so, there are plenty of potential
applications: replacing the operad ${\operatorname{Com}}$ by the
$\infty$-operad $E_{\infty}$ of commutative algebras up to infinity could give
an insight into the well-defined notion of derived algebraic geometry [34].
Using a replacement for ${\operatorname{Lie}}$ could recapture notions from
the geometry of Lie algebras of Francis and Gaitsgory, and from works of
Harpaz, Nuiten and Prasma [14, 20]. Replacing ${\operatorname{Ass}}$ by the
$\infty$-operad $A_{\infty}$ should recapture the theory of
$A_{\infty}$-geometry of Kontsevich and Soibelman [25].

  5. (v)

In Sections 4 we discussed briefly the functoriality of the two constructions.
In particular, we mentioned how every operad morphism provides strong/strict
tangent morphisms relating to each construction. This fact should play an
important role in better understanding the link between operad theory and
tangent category theory.

So there are many potential interesting paths to take for future work
regarding operads and tangent categories.

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