Poisson Lie Group
Can group looks like 'mechanical'?
The algebra of tensor representation of Lie group is mechanical, and this called Poisson Lie group.
Hopf algebra
Definition: An (unital associative)algebra is a linear space $A$ together with two maps
$m\ :\ A \otimes A \rightarrow A$
$\eta\ :\ \mathbf{C} \rightarrow A$
such that
1. $m$ and $\eta$ are linear
2. $m(m\otimes 1) = m(1\otimes m)$ (associativity)
3. $m(1\otimes \eta) = m(\eta \otimes 1) = id$ (unit)
Definition: Let $(A, m, \eta)$ be an algebra, $V$ a linear space and $\rho$ a map from $A$ to the space of linear operators in $V$. $(V, \rho)$ is called a representation of $A$ if
1. $\rho$ is linear
2. $\rho (xy) = \rho(x) \rho(y)$
Tensor product of two representation $(\psi_1, V_1)$ and $(\psi_2, V_2)$ action of $A$ an $V_1 \otimes V_2$ is $\Psi$ with a map $\Delta : A \rightarrow A\otimes A$
$\Psi = (\psi_1 \otimes \psi_2) \Delta$
To $\Psi$ be linear, satisfy homomorphism property and representations $V_1 \otimes V_2)\otimes V_3$ and $V_1 \otimes (V_2 \otimes V_3)$ to be equal, $\Delta$ follows these conditions, and it called co-associativity.
1. $\Delta$ is linear
2. $\Delta (ab) = \Delta(a) \Delta(b)$ (homomorphism)
3. $(\Delta \otimes id) \Delta = (id \times \Delta) \Delta$
This commutativity of diagram is called co-associativity.
Definition: A co-unit is a map $\epsilon: A \rightarrow \mathbf{C}$ such that $(1\otimes \epsilon) \Delta = (\epsilon \otimes 1) \Delta = id$
Example: Commutative Hopf algebra
Let $G$ be a compact topological group. Consider the space of continuous functions on $G$ denoted by $C(G)$ together with the following maps:
$\quad \bullet\ (f.h)(g) = f(g)h(g)$
$\quad \bullet\ \Delta (f)(g_1 \otimes g_2) = f(g_1 g_2)$
$\quad \bullet\ \eta(x) = x1$ where $1(g) = 1$ for all $g\in G$
$\quad \bullet\ \epsilon(f) = f(e)$ where $e$ is the unit element of $G$
$\quad \bullet\ S(f)(g) = f(g^{-1})$
where $g_1, g_2, g \in G$, $x\in \mathbf{C}$ and $f,h\in C(G)$.
Then $C(G)$ together with these maps is a Hopf algebra.
Example: Universal enveloping algebra(co-commutative Hopf algebra)
Let $L$ be a Lie algebra and $\mathcal{U}(L)$ its universal enveloping algebra, then $\mathcal{U}(L)$ becomes a Hopf algebra if we define
$\quad \bullet\ $The multiplication is the ordinary multiplication in $\mathcal{U}(L)$
$\quad \bullet\ \Delta (x) = x \otimes 1 + 1 \otimes x$
$\quad \bullet\ \eta(\alpha) = \alpha 1$
$\quad \bullet\ \epsilon (1) = 1$ and zero on all other elements
$\quad \bullet\ S(x) = -x$
where $x$ is an element of $L$.
There is equality $\Delta = \tau \circ \Delta$ where $\tau$ is again the flip operator, so it is called co-commutative.
Like above example, dual of Hopf algebra is also Hopf algebra.
Let $(A, m ,\Delta, \eta, \epsilon, S)$ is a Hopf algebra and that $A^*$ is its dual space, then using the structure maps of $A$ we defined the structure maps $(m^*, \Delta^*, \eta^*, \epsilon^*, S^*)$ on $A^*$ as follows:
$\quad \bullet\ \left\langle m^* (f\otimes g), x\right\rangle = \left\langle f\otimes g, \Delta (x) \right\rangle$
$\quad \bullet\ \left\langle \Delta^* (f), x\otimes y \right\rangle = \left\langle f, xy \right\rangle$
$\quad \bullet\ \left\langle \mu^* (\alpha), x\right\rangle = \alpha . \epsilon (x)$
$\quad \bullet\ \epsilon^*(f) = \left\langle f, 1\right\rangle$
$\quad \bullet\ \left\langle S^* (f), x\right\rangle = \left\langle f, S(x)\right\rangle$
where $f$ and $g$ are elements of $A^*$ and $x,y$ are elements of $A$. The brakets $\left\langle .,.\right\rangle$ denote the dual contraction between $A^*$ and $A$, and $\left\langle f \otimes g, x\otimes y\right\rangle = \left\langle f, x \right\rangle \left\langle g, y \right\rangle$.
Let's find duality between $\mathcal{U}(L)$ and $C^\infty(G)$.
Consider the map
$\rho\ :\ L \rightarrow \mbox{End}_\mathbf{C}(C^\infty(G))$
defined by
$(\rho (X) \phi)(g) = \frac{d}{dt} (\phi (e^{tX}g))|_{t=0}$
for $X\in L$, $\phi \in C^infty(G)$, $g\in G$. (derivative in $g$ of $\phi$ in the direction $X$)
This map extends uniquely to a homomorphism
$\rho: \mathcal{U}(L) \rightarrow \mbox{End}_\mathbf{C}(C^\infty(G))$
by definition of $\mathcal{U}(L)$. Also consider the right action of $G$ on $C^\infty(G)$
$R_g\ :\ C^\infty(G) \rightarrow C^\infty(G)$
defined by $R_g(\phi)(g') = \phi(g'g)$.
Lemma: The map $\rho$ has the following properties.
1. $R_g \circ \rho(a) = \rho(a) \circ R_g$ (right invariance)
2. $\rho(X)(\phi . \psi) = (\rho (X)\phi ).\psi + \phi .(\rho(X)\psi)$ (derivation property)
for all $a\in \mathcal{U}(L)$, $X\in L$ and $\phi,\psi \in C^\infty(G)$.
proof:
$(\rho(X)\circ R_g)(\phi)(g') = \frac{d}{dt} \phi(e^{tX} g'g) |_{t=0} = (R_g\circ \rho(X))(\phi)(g')$ for $X\in L$.
Second is Leibniz rule.
Let's define the pairing
$\left\langle .,.\right\rangle : C^\infty(G) \times \mathcal{U}(L) \rightarrow \mathbf{C}$
by $\left\langle \phi, a\right\rangle = (\rho (a) \phi)(e)$ where $e$ is the unit element of $G$, $a\in \mathcal{U}(L)$ and $\phi\in C^\infty(G)$.
Theorem: The map $C^\infty (G) \rightarrow (\mathcal{U}(L))^*$ defined by $\phi\rightarrow \left\langle \phi, .\right\rangle$ us an embedding.
proof: Goal is to prove that this map is injective. Suppose $\left\langle \phi_1, a\right\rangle = \left\langle \phi_2, a\right\rangle$ for all $a\in \mathcal{U}(L)$. Then $0=\left\langle \phi_1 - \phi_2, a\right\rangle = (\rho (a)(\phi_1 - \phi_2))(e)$.
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Poisson Structure
Definition: A Poisson algebra is a commutative algebra $(A, m, \eta)$ together with a map
$\{ .,. \}\ :\ A \times A \rightarrow A$
such that
1. $A$ is a Lie algebra with respect to $\{ .,. \}$. (Antisymmtery, Jacobi identity)
2. $\{ab, c\} = a\{ b,c\} + \{ a, c\} b$. (Leibniz identity)
Space of smooth functions on a symplectic manifold is a Poisson algebra.
For tensor product language, $\{ .,. \}$ induce a map $\gamma : A\otimes A \rightarrow A$.
1. $\gamma \circ \tau = -\gamma$ (anti-symmetry)
$\gamma (1\otimes \gamma) (1\otimes 1 \otimes 1 + (1 \otimes \tau)(\tau \otimes 1) + (\tau \otimes 1)(1 \otimes \tau)) = 0$ (Jacobi-identity)
2. $\gamma (m\otimes 1) = m(1\otimes \gamma)(1\otimes 1 \otimes 1 + \tau \otimes 1)$
Definition: Let $(A, m_A, \{ .,.\}_A)$ and $(B, m_B, \{ .,.\}_B )$ be Poisson algebras. A Poisson algebra homomorphism is a linear map $f$ from $A$ to $B$ such that
1. $f(a.b) = f(a)f(b)$
2. $f(\{a,b\}_A) = \{f(a),f(b)\}_B$
where $a$ and $b$ are arbitrary elements of $A$.
For more clear expression with $m$ and $\gamma$,
1. $f\circ m_A = m_B \circ (f \otimes f)$
2. $f\circ \gamma_A = \gamma_B \circ (f\otimes f)$
Theorem: Tensor product of space $A\otimes B$ of the two Poisson algebra became also Poisson algebra.
proof: Multiply two elements of $A\otimes B$ is $(a\otimes b).(c\otimes d) = (a.b)\otimes (c.d)$ where $a,c\in A$ and $b,d\in B$, or $m_{A\otimes B} = (m_A \otimes m_B)(1\otimes \tau \otimes 1)$ where $\tau$ is flip operator.
Poisson structure preserve since $\{a\otimes b, c\otimes d\}_{A\otimes B} = \{a, c\}\otimes bd + ac\otimes \{b,d\}$ or $\gamma_{A\otimes B} = (\gamma_A \otimes m_B + m_A \otimes \gamma_B)(1\otimes \tau \otimes 1)$
If tensor product to be a Poisson algebra homomorphism, $\Delta$ must satisfy condition
$\{ \Delta (a), \Delta (b)\}_{A\otimes A} = \Delta (\{a, b\}_A)$ or $\Delta_{A\otimes A} \circ (\Delta \otimes \Delta) = \Delta\circ \gamma_A$
and $A$ is called a Poisson bi-algebra.
Definition: A co-Poisson bi-algebra is a co-commutative bi-algebra $(A,m,\Delta,\eta,\epsilon)$ together with a map $\delta : A\rightarrow A \otimes A$ such that
1. $\tau \circ \delta = -\delta$ (co-antisymmetry)
2. $(1\otimes 1 \otimes 1 + (1\otimes \tau )(\tau \otimes 1) + (\tau \otimes 1)(1 \otimes \tau))(1\otimes \delta)\delta = 0$ (co-Jacobi identity)
3. $(\Delta \otimes 1)\delta = (1\otimes 1\otimes 1\otimes + \tau \otimes 1)(1\otimes \delta)\Delta$ (co-Leigniz rule)
4. $(m\otimes m)\circ \delta_{A\otimes A} = \delta \circ m$ ($m$ is co-Poisson homomorphism.)
where $\delta_{A\otimes A} = (1\otimes \tau \otimes 1)(\delta \otimes \Delta + \Delta \otimes \delta)$ is the co-Poisson structure naturally associated to the tensor product space.
Poisson-Lie groups and Lie bi-algebras
Definition: A Lie group $G$ is called a Poisson Lie group if the space of smooth functions on $G$ is a Poisson Hopf algebra.
Lemma: Let $\{X_\mu\}^{dim(G)}_{\mu=1}$ be a set of right invariant vectorfields on $G$ (i.e. if $R_g(g') = g'g$, then $X_\mu |_g = (R_g)_* X_\mu |_e$ where $(R_g)_*$ is the derivative of $R_g$, and $e$ is the unit element of $G$) such that $\{X_\mu |_g\}$ is a basis in $T_g G$ for all $g\in G$. Then a Poisson bracket on $C^\infty (G)$ can be written as $\{\phi,\psi\} = \sum_{\mu\nu} \eta^{\mu\nu} (g) X_\mu |_g (\phi) X_\nu |_g (\psi)$ where $g\in G$.
proof: Because $\{\phi, .\} : C^\infty (G) \rightarrow C^\infty (G)$ is a derivation, so it's vectorfield. Then basis for tangent space $T_g G$ for every $g\in G$ is $\{X_\mu |_g\}$, so we can write $\{ \phi ,.\} (g) = \sum_\mu \gamma^\mu (g) X_\mu |_g$. Apply also for $\{.,\psi\}$.
This can express differently: $\{\phi,\psi\}(g) = \sum_{\mu\nu} \eta^{\mu\nu} (g) X_\mu|_g (\phi) X_\nu |_g (\psi) = \sum_{\mu\nu} \eta^{\mu\nu} (g) (d\phi |_g \otimes d\psi |_g) (X_\mu|_g \otimes X_\nu|_g) = \eta(g) (d\phi |_g \otimes d\psi |_g)$ where $\eta : G \rightarrow L\otimes L$ is defined by ($L$ is the Lie algebra of $G$, $T_e G$.) $g \longmapsto \eta(g) = \sum_{\mu\nu} \eta^{\mu\nu} (g) X_\mu \otimes X_\nu$ and also $d\phi |_g (X_\mu) \equiv d\phi |_g (X_\mu |_g)$.
Let $C^n (G; L)$ be the space of maps $\lambda : G \times \cdots \times \rightarrow L\otimes L$.
We can turn the sepquence $\{C^n(G;L)\}^\infty_{n=0}$ into a complex by defining the coboundary operator
$\delta_G: C^n(G;L)\rightarrow C^{n+1} (G;L)$
as follows $[\delta_G \lambda] (g_1, \cdots, g_{n+1}) = $
$g_1.\lambda (g_2, \cdots, g_{n+1}) + \sum^n_{i=1}(-1)^i\lambda (g_1, \cdots, g_i g_{i+1}, \cdots, g_{n+1}) + (-1)^{n+1} \lambda(g_1, \cdots, g_n)$, where $g_i\in G$ and $\lambda \in C^n (G;L)$.
The action of $G$ on $L\otimes L$ is defined by
$g.(X\otimes Y) = Ad_g X\otimes Ad_g Y \equiv Ad^{\otimes 2}_g (X\otimes Y)$
where $Ad$ denotes the adjoint action and $X,Y\in L$.
Also we can check $\delta^2_G = 0$.
Theorem: Let $\eta$ be the map associated to the Poisson structure on $C^\infty(G)$ via the relation $\{\phi, \psi\}(g) = \eta(g)(d\phi |_g \otimes d\psi |_g)$. Then the compatibility relation $\{\Delta (a), \Delta (b) \}_{A\otimes A} = \Delta (\{a, b\}_A)$ between the Poisson bracket and the Hopf algebra structure on $C^\infty(G)$ is equivalent to the cocycle condition on $\eta$, i.e.
$\delta_G \eta = g_1._\eta(g_2) - \eta(g_1g_2) +\eta(g_1) = 0$
proof: Let write $\Delta (\phi) = \sum_i \phi^{(1)}_i \otimes \phi^{(2)}_i$ then by definition of the co-product on $C^\infty (G)$, $\Delta \phi(g_1, g_2) = \phi(g_1 g_2) = \sum_i \phi^{(1)}_i (g_1) \phi^{(2)}_i (g_2)$.
By definition of the Poisson bracket on the tensor product space $C^\infty (G)\times C^\infty (G)$, $\{\Delta (\phi), \Delta (\psi)\} = \sum_{ij} \{\phi^{(1)}_i, \psi^{(1)}_j\}\otimes \phi^{(2)}_i \psi^{(2)}_j + \phi^{(1)}_i \psi^{(1)}_j \otimes \{\phi^{(2)}_i,\psi^{(2)}_j\}$, we find
$\{\Delta (\phi),\Delta(\psi)\}(g_1,g_2) = \sum_{\mu\nu}\sum_{ij}(\eta^{\mu\nu}(g_1) X_\mu|_{g_1}(\phi^{(1)}_i) X_\nu|_{g_1}(\psi^{(1)}_j)\phi^{(2)}_i(g_2)\psi^{(2)}_j(g_2) + \phi^{(1)}_i(g_1)\psi^{(1)}_j(g_1)(\eta^{\mu\nu}(g_2) X_\mu|_{g_2}(\phi^{(2)}_i)X_\nu|_{g_2} (\psi^{(2)}_j))$.
Also find $\sum_i X_\mu|_{g_1} (\phi^{(1)}_i)\phi^{(2)}_i(g_2) = \frac{d}{dt} \sum_i \phi^{(1)}_i (e^{iX_\mu}g_1)\phi^{(2)}_i(g_2)|_{t=0} = \frac{d}{dt}\phi(e^{iX_\mu} g_1g_2)|_{t=0} = d\phi|_{g_1g_2} (X_\mu)$.
Similarly, $\sum_i \phi^{(1)}_i(g_1)X_\mu|_{g_2} (\phi^{(2)}_i) = d\phi|_{g_1g_2} (Ad_{g_1}X_\mu)$.
Use above result, $\{\Delta(\phi),\Delta(\psi)\}(g_1,g_2) = (d\phi|_{g_1g_2}\otimes d\psi|_{g_1g_2})(\eta(g_1) + g_1.\eta(g_2)) $.
Use definition, $\Delta \{\phi,\psi\}(g_1,g_2) = (d\phi|_{g_1g_2}\otimes d\psi|_{g_1g_2}).\eta(g_1g_2)$.
Associated to the map $\eta$ we define $\phi_\eta : L\rightarrow L\otimes L$ by $\phi (X) = \frac{d}{dt} \eta(e^{tX})|_{t=0}$.
Theorem: Let $\eta$ be the map associated to the Poisson structure on the Poisson Lie group G and let $\phi_\eta$ be detinde by above formular, then
1. $\phi_\eta$ is co-antisymmetric.
2. $\phi_\eta$ satisfies the co-Jacobi identity.
3. $\phi_\eta([X,Y]) = X.\phi_\eta(Y) - Y.\phi_\eta(X)$
where $L$ acts on $L\otimes L$ via $X.(Y\otimes Z) = [X,Y]\otimes Z + Y \otimes [X,Z]$ which is the infinitesimal version of the action of $G$ on $L\otimes L$.
proof: Anti-symmetry of the Poisson bracket gives $\eta(d\phi\otimes d\psi) = \sum \eta^{\mu\nu} (g) X_\mu(\phi) X_\nu(\psi) = -\sum \eta^{\mu\nu}(g) X_\mu(\psi)X_\nu(\phi) = -(\tau\circ\eta)(d\phi\otimes d\psi)$ so indeed we fined $\eta = -\tau\circ\eta$. Use definition of $\phi_\eta$, $\tau\circ\phi_\eta = -\phi_\eta$. co-Jacobi identity is similar.
From co-cycle condition $\eta(e)=0$ with $g_1 = e$, $0=\partial_t \eta(e^{tX}e^{-tX})|_{t=0} = \partial_t \eta(e^{tX})|_{t=0} + \partial_t ((Ad^{\otimes 2}_{e^{tX}}\eta(e^{-tX}))|_{t=0} = \phi_\eta(X) +\phi_\eta(-X)$.
Then $\phi_\eta([X,Y]) = \frac{d}{ds}\frac{d}{dt}\eta(e^{sX}e^{tY}e^{-sX}|_{t=0} = \frac{d}{ds}\frac{d}{dt}(\eta(e^{tX})+(Ad^{\otimes 2}_{e^{sX}})\eta(e^{tY})+Ad^{\otimes 2}){e^{tX}} Ad^{\otimes 2}_{e^{tY}} \eta(e^{-sX})|_{s,t=0} = \frac{d}{ds}(Ad^{\otimes 2}_{e^{sX}})|_{s=0} \phi_\eta(Y) + \frac{d}{dt}(Ad^{\otimes 2}_{e^{tY}})|_{t=0}\phi_\eta(-X) = Ad^{\otimes 2}_X \phi(Y) + Ad^{\otimes 2}_Y \phi(-X) = X.\phi_\eta(Y) - Y.\phi_\eta(X)$
This called Lie bi-algebra.
Most simple way to satisfy the co-cycle condition $\delta_G\eta = 0$ is coboundary $\eta = \delta_G r$ for some $r\in L\otimes L$.
For that case, $[\delta_Gr](g) = r - g.r$ so $\phi(X) = \frac{d}{dt}\eta(e^{iX})|+{t=0} = \frac{d}{dt}(r-e^{tX}.r)|_{t=0} = -X.r$.
When write $r = r^{\mu\nu} X_\mu \otimes X_\nu$, then $\phi(X) = -r^{\mu\nu}([X,X_\mu]\otimes X_\nu + X_\mu \otimes [X,X_\nu]) = [r,1\otimes X+X\otimes 1]$.
To consider co-Jacobi identities and co-antisymmetry,
Theorem: Let $L$ be a Lie algebra and $r$ an element of $L\otimes L$. Choose and arbitrary basis $\{X_\mu\}$ in $L$ and write $r=r^{\mu\nu} X_\mu \otimes X_\nu$. Also detine
$r_+ = \frac{1}{2} (r^{\mu\nu} + r^{\nu\mu})X_\mu\otimes X_\nu$
$r_- = \frac{1}{2} (r^{\mu\nu} - r^{\nu\mu})X_\mu\otimes X_\nu$
$r_{12} = r^{\mu\nu} X_\mu \otimes X_\nu \otimes 1$
$r_{13} = r^{\mu\nu} X_\mu \otimes 1 \otimes X_\nu$
$r_{23} = r^{\mu\nu} 1 \otimes X_\mu \otimes X_\nu$
Then the map $\phi : L\rightarrow L\otimes L$ defined by $\phi(x) = [r,X\otimes 1 + 1 \otimes X]$ turns $L$ into a Lie bi-algebra if and only if
1. $r_+$ is ad-invariant, i.e. $(Ad_g \otimes Ad_g)r_+ = r_+$
2. $B = [r_{12}, r_{13}] + [r_{13},r_{23}] + [r_{12},r_{23}]$ is ad-invariant, i.e. $(Ad_g \otimes Ad_g \otimes Ad_g)B = 0$ for all $g\in G$. $B$ is called the schouten bracket of $r$ with itself.(? 0맞나?)
proof: use co-antisymmetry and co-Jacobi identities for $\phi$.
$Ad^{\otimes 3}_g B = 0$ is called modified classical Yang-Baxter equation, $B=0$ is called classical Yang-Baxter equation.
Example: Let $L = sl_N$ and let $r$ be given by $r = C - \sum_{i<j} e_{ij} \wedge e_{ji}$ where $C = \sum_{i\ne j} e_{ij} \otimes e_{ji} + \sum^{N-1}_{\mu\nu=1} K^{\mu\nu} H_\mu\otimes H_\nu$ and $(e_{ij})_{kl} = \delta_{ik}\delta_{jl}$. $H_\mu$ are the standard generators of the Cartan subalgebra. $K^{\mu\nu}$ is the inverse of Cartan matrix.
In the special case of $sl_2$ this reduces to $r = \frac{1}{2} H\otimes H + 2E\otimes F$ such that $\phi$ becomes $\phi(H) = 0$, $\phi(E)=\frac{1}{2} E\wedge H$, $\phi(F) = \frac{1}{2} F\wedge H$.
Since $\eta(g) = r - Ad_g r$, $\eta^{\mu\nu} (g) X_\mu|_g \otimes X_\nu|_g = r^{\mu\nu}(X_\mu|_g\otimes X_\nu|_g - Ad_g X_\mu |_g \otimes Ad_g X_\nu |_g)$.
Use $X_\mu|_g=(R_g)_*X_\mu|_e$ and $Ad_g = (L_g)_*(R^{-1}_g)_*$, find $Ad_g X_\mu |_g = (L_g)_* X_\mu |_e$.
Denote $(R_g)_* X_\mu|_e(\phi)$ by $\partial_\mu \phi$ and $(L_g)_* X_\mu|_e$ by $\partial'_\mu \phi$ we get $\{f,g\} = r^{\mu\nu} (\partial_\mu f \partial_\nu g - \partial'_\nu f \partial'_\nu g)$.
Universal enveloping algebra of a Poisson Lie group $G$ is a co-Poisson Hopf algebra. Denote the co-Poisson structure on $\mathcal{U}(L)$ by $\delta$.
Theorem: The restriction of $\delta$ to $L \subset \mathcal{U}(L)$ is equal to $\phi_\eta$.
proof: Denote Poisson structure by $\gamma$, i.e. $\{\phi,\psi\} = \gamma (\phi\otimes\psi)$. The map $\delta$ is defined by the relation $\left\langle \gamma(\phi\otimes\psi),a\right\rangle = \left\langle \phi\otimes\psi, \delta(a)\right\rangle$ where $\phi,\psi \in C^\infty (G)$, $a\in \mathcal{U}(L)$ and $\left\langle .,.\right\rangle$ denotes the duality between $C^\infty(G)$ and $\mathcal{U}(L)$. Denoting $X_\mu\phi$ by $\partial_\mu\phi$ by the left hand side of this equation is euqal to $\sum_{\mu\nu} \left\langle \eta^{\mu\nu} \partial_\mu \phi \partial_\nu\psi, X\right\rangle = \rho(X) (\eta^{\mu\nu} \partial_\mu\phi\partial_\nu\psi)(e) = \frac{d}{dt} \eta^{\mu\nu}(e^{tX})\partial_\mu\phi(e^{tX})\partial_\nu\psi(e^{tX})|_{t=0} = \frac{d}{dt} \eta^{\mu\nu}(e^{tX})|_{t=0} \partial_\mu\phi(e)\partial_\nu\psi(e) = \phi^{\mu\nu}_\eta (X) X_\mu|_e(\phi) X_\nu|_e(\psi)$ for $X\in L\subset \mathcal{U} (L)$. Since $\left\langle \phi, X\right\rangle = \frac{d}{dt} \phi(e^{tX})|_{t=0} = X|_e(\phi) = d\phi|_e(X)$, find $\left\langle \phi \otimes\psi, \delta(x)\right\rangle = \left\langle(d \phi|_e \otimes d\psi|_e)\delta(X)\right\rangle$ for the right hand side.
뭔소리지 이게
We will use this for quantum group.
Reference
T.Tjin - An introduction to quantized Lie groups and algebras
A Guide to Quantum Groups