Tangent space of the pre-image of identity via the Lie product Let $G$ be a Lie group and denote by $\mu:G\times G\rightarrow G$ its product. I am trying to understand the submanifold $\mu^{-1}(\{e\})=\{(g,g^{-1}):g\in G\}$ in the Lie group product $G\times G$. More precisely, I would like to characterize the tangent space
$$ T_{(g, h)}(\mu^{-1}(\{e\}))\subset T_gG\oplus T_{h}G .$$
I have tried to use the fact that $\mu$ is a submersion and so such tangent space is given by $\ker d\mu_{(g,h)}$. Therefore, if $$(X,Y)\in T_{(g, h)}(\mu^{-1}(\{e\})),$$ then, using that $d\mu_{(g,h)}(X,Y)=d(L_g)_h Y+d(R_h)_gX=0$, we get
$$
X=-d(R_{h^{-1}}\circ L_g)_h Y,
$$
and since $h=g^{-1}$, it follows that
$$
X=-d(R_g\circ L_g)_{g^{-1}}Y,$$
but I can't see if there is a better characterization for this problem. Any ideas would be appreciated!
 A: There are the canonical projection maps $p,q: G\times G \rightarrow G$ defined by
$$p(g,h):=g, q(g,h):=h.$$
Let $Z:=\mu^{-1}(e) \subseteq G\times G$.
There is a canonical map
$$s: G \rightarrow Z \subseteq G \times G$$
defined by $s(g):=(g,g^{-1})$. We get induced map $p,q: Z \rightarrow G$ defined by
$p(g,g^{-1}):=g, q(g,g^{-1}):=g^{-1}$. It follows
$$sp(g,g^{-1})=s(g)=(g,g^{-1})$$
and
$$ps(g)=p(g,g^{-1})=g$$
hence $s: G \rightarrow Z$ seems to be an "isomorphism of manifolds". The set $Z$ is not a subgroup of $G\times G$ since
$$(g,g^{-1})(h,h^{-1}):=(gh, g^{-1}h^{-1}) \notin Z$$
and the map $s$ is not a map of Lie groups. If you give $Z$ the following product
$$(g,g^{-1})(h,h^{-1}):=(gh, h^{-1}g^{-1})=(gh,(gh)^{-1})$$
it follows $s$ is an isomorphism of Lie groups.
Question: "More precisely, I would like to characterize the tangent space
$T_{(g,h)}(μ^{−1}({e})) \subseteq T_g(G)\oplus T_h(G)$."
Answer: It seems to me the tangent space of $Z$ at $x:=(g,g^{-1})$ should be as follows
$$T_x(Z) \cong T_g(G).$$
This is beacause $s$ is an isomorphism and $p(x)=g$.
