How to show that $t \subset g$ is a Lie sub-bialgebra if and only if $t^{\perp} \subset g^*$ is an ideal? Let $g$ be a Lie bialgebra. That is, $g$ is a Lie algebra, there is a linear map $\delta: g \to g \otimes g$ such that $\delta^*: g^* \otimes g^* \to g^*$ is a Lie bracket on $g^*$, and 
$$
{\displaystyle \delta ([X,Y])=\left(\operatorname {ad} _{X}\otimes 1+1\otimes \operatorname {ad} _{X}\right)\delta (Y)-\left(\operatorname {ad} _{Y}\otimes 1+1\otimes \operatorname {ad} _{Y}\right)}.
$$
Let $t \subset g$ and $t^{\perp} = \{f \in g^*: f(s)=0, \forall s \in t\}$. I am trying to show that $t$ is a Lie sub-bialgebra of $g$ if and only if $t^{\perp} \subset g^*$ is an ideal.
Suppose that $t$ is a Lie sub-bialgebra of $g$. Then $[s_1, s_2] \in t$, for all $s_1, s_2 \in t$ and $\delta(s) \in t \otimes t$ for all $s \in t$. Let $f \in t^\perp$ and $g \in g^*$. I need to show that for all $s \in t$, $[f, g](s) = 0$. How to prove this? Thank you very much.
 A: You have a Lie subalgebra $\mathfrak t\subset\mathfrak g$ and you want to show that $\mathfrak t\subset\mathfrak g$ is a Lie sub-bialgebra, i.e. that $\delta(\mathfrak t)\subset \mathfrak t\otimes\mathfrak t$, iff $\mathfrak t^\perp\subset \mathfrak g$ is an ideal. For this just observe that 
$$(\mathfrak t\otimes\mathfrak t)^\perp=\mathfrak t^\perp\otimes\mathfrak g^*+\mathfrak g^*\otimes\mathfrak t^\perp\subset\mathfrak g^*\otimes\mathfrak g^*$$
and thus $\mathfrak t$ is a Lie sub-bialgebra iff
$$\langle \delta(t),\alpha\otimes\beta\rangle=0$$
for every $t\in\mathfrak t$, $\alpha\in\mathfrak g^*$, $\beta\in\mathfrak t^\perp$ (using that $\delta$ is skew-symmetric). By the definition of the Lie bracket on $\mathfrak g^*$ we have $\langle \delta(t),\alpha\otimes\beta\rangle=\langle t,[\alpha,\beta]\rangle$, and thus the condition is: for every $\alpha\in\mathfrak g^*$, $\beta\in\mathfrak t^\perp$, we have $[\alpha,\beta]\in\mathfrak t^\perp$, i.e. $\mathfrak t^\perp\subset\mathfrak g^*$ is an ideal.
