Let $G$ be Lie group with identity $e$ and dimension $n$. Let $T_eG$ be the Lie algebra of $G$. Let $Z_e$ $\in T_eG$ be the identity of $T_eG$.

Consider the exponential map $\exp: T_eG \to G$. We have $\exp(Z_e) = e$. Therefore, the differential of $\exp$ at $Z_e$ is $\exp_{\{*, Z_e\}}:$ $T_{Z_e}(T_eG)$ $\to$ $T_eG$.

There exists a unique canonical isomorphism $\gamma: T_{Z_e}(T_eG) \to T_eG$ that allows us to identify the double tangent space $T_{Z_e}(T_eG)$ with the Lie algebra $T_eG$. This identification apparently allows us to say

'$\exp_{\{*, Z_e\}}: T_eG \to T_eG$' is the identity map. $\tag{Statement A}$

Questions: 1. (I have an idea on how to answer the first question, and I ask about this in the second question) What does Statement A mean exactly? 2. Is it correct that Statement A is equivalent to Statement B as follows? 3. If yes, then is the differential (with the original domain) $\exp_{\{*, Z_e\}}:$ $T_{Z_e}(T_eG)$ $\to$ $T_eG$ actually $\gamma$ itself?

$\exp_{\{*, Z_e\}} \circ \gamma^{-1}:$ $T_eG \to T_eG$ is the identity map. $\tag{Statement B}$

There is a similar or related question I recently asked: Showing differential of left multiplication in real general linear group is left multiplication using left multiplication in $\mathbb R^{n \times n}$

Thanks in advance!

Some notes:

  1. I'm aware that and aware how $T_eG$ is itself a Lie group (and thus a smooth manifold) besides an $\mathbb R$-vector space with finite dimension. This has been asked before, and this is not what I'm asking.

  2. It has been previously asked before how to prove Statement A. This is not what I am asking. I am asking what Statement A means if Statement A is not equivalent to Statement B and asking if $\exp_{\{*, Z_e\}}$ (the original) is actually $\gamma$.

  • $\begingroup$ The underlying question here is not specific to Lie groups. For any real or complex vector space $\Bbb V$ and any vector $v \in \Bbb V$ there is a canonical isomorphism $T_v \Bbb V \to \Bbb V$. $\endgroup$ – Travis Willse Nov 12 at 22:17
  • $\begingroup$ @TravisWillse To be absolutely clear, I understand that there is a (unique I think) canonical $\mathbb R$-vector space isomorphism and diffeomorphism (and Lie group isomorphism) $\delta$ between the tangent space $T_pM$ and the Lie group and $n$- dimensional $\mathbb R$-vector space $\mathbb R^n$, for $M$ a smooth manifold with dimension $n$, whether or not $M$ is a vector space, and for $p \in M$. $\endgroup$ – Ekhin Taylor R. Wilson Nov 13 at 8:14
  • $\begingroup$ @TravisWillse I edited my post to be clearer about the questions. Can you please explain what your answers to the questions specifically are? $\endgroup$ – Ekhin Taylor R. Wilson Nov 13 at 8:30

This is exactly what it means. Once we identify $T_0(T_eG)$ with $T_eG$, in the standard way, with $v(f)|_u=\frac{d}{dt}(f(u+tv))|_{t=0}$ we may write $d_0\exp$ as a map from $T_eG$ to $T_eG$.

As a side note on terminology, "double tangent space" frequently refers to the fibers of the double tangent bundle $T_v(TM)$, which is very different from the tangent space of the tangent space $T_v(T_pM)$.

  • $\begingroup$ Kajelad, I did some editing to hopefully be clearer to Travis Willse. To clarify, your answers to the second and third questions are affirmative while your answer to the first question is 'The guess asked about in the second question is correct'? $\endgroup$ – Ekhin Taylor R. Wilson Nov 13 at 8:35
  • 1
    $\begingroup$ Yes, more or less. When we "identify" spaces in this manner we are implicitly composing with whatever canonical isomorphisms we use to set up the identification; there is always only one way to do this. These isomorphisms can always be written out explicitly (as you have), though in many contexts in differential geometry it is needless to do so. $\endgroup$ – Kajelad Nov 13 at 8:48

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