1

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}=T_e G$. The Kirillov-Kostantstructure Poisson structure on $\mathfrak{g}^*$ is defined as $\{ f,g\} (p)=p([f_{*p},g_{*p}])$ where $f,g\in C^{\infty}(\mathfrak{g}^*)$ and $f_{*p}$ denotes the differential of $f$ at $p\in \mathfrak{g}^*$.

My main question is that why $f_{*p}\in \mathfrak{g}$?

Indeed, $f_{*p}:T_p\mathfrak{g}^*\to T_{f(p)}\mathbb{R}$. I guess that $T_p\mathfrak{g}^*\cong \mathfrak{g}^*$ and $T_{f(p)}\mathbb{R}\cong \mathbb{R}$. If so, then $f_{*p}\in (\mathfrak{g}^*)^*$. I also guess $(\mathfrak{g}^*)^*\cong \mathfrak{g}$, so $f_{*p}\in \mathfrak{g}$. But

(1) how can I define the isomorphism $T_p\mathfrak{g}^*\cong \mathfrak{g}^*$?

(2) how can I define the isomorphism $(\mathfrak{g}^*)^*\cong \mathfrak{g}$?

(3) How can I directly see $f_{*p}$ as an element of $\mathfrak{g}$?

Sorry if I ask a lot of questions, but I really like to know and see details to understand them well. Thank you very much in advance.

Mahtab
  • 727

1 Answers1

1

Your guesses are all correct.

  1. If $V$ is a vector space (like $V = \mathfrak{g}$ or $V = \mathbb{R}$), then $T_p V \cong V$ for any $p \in V$. Basically this is because vectors in a vector space can be “moved around”, so you can choose to “attach” them to the point $p$ if you like.

  2. Any finite-dimensional vector space $V$ (like $V = \mathfrak{g}$) is naturally isomorphic to its double dual $V^{**}$. See this question, for example.

  3. If you pick a basis for $\mathfrak{g}$ then you get a corresponding dual basis for $\mathfrak{g}^*$ (as is also explained in the question that I linked to above). Let the coordinates with respect to that dual basis be called $(x_1,\dots,x_n)$, so that a function $f \in C^\infty(\mathfrak{g}^*)$ would be written as $f(x_1,\dots,x_n)$. Then $f_{*p}$ is simply identified with the element of $\mathfrak{g}$ whose coordinates with respect to the first basis are $$ \Bigl(\tfrac{\partial f}{\partial x_1}(p), \dots, \tfrac{\partial f}{\partial x_n}(p) \Bigr) . $$

Hans Lundmark
  • 55,038