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I am an undergraduate unfamiliar with Lie theory. I am using some Lie theory in an operational manner for a physics-related project. However, I would like to get a better understanding of what I am using.

Let $G$ be a matrix Lie group and let $T_g G$ denote the tangent space at $g \in G$. Suppose we know the directional derivatives of $G$ at the identity $\text{id} \in G$ and denote them by the set $\{\Lambda_n\}$.

It seems that one can use the pushforward $(L_g)_* : T_{\text{id}}G \rightarrow T_gG$ defined by $(L_g)_*(\Lambda_in) = g \cdot \Lambda_n \equiv \delta_n g$, where $\cdot$ is matrix multiplication, to send directional derivatives at the identity to directional derivatives at $g \in G$. Is this accurate?

If so, how do I compute $\delta_n g^{-1}$? I naively thought it would just be $g^{-1} \Lambda_n$. However, I was told that it is actually $-\Lambda_n g^{-1}$ with the concept of a pullback mentioned.

Any resources on this topic would also be much appreciated!

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When you write that $\{\Lambda_n\}$ are the directional derivatives of $G$ at $\mathrm{id}$, I guess it means that $\dim(G) = n$ and $(\Lambda_1,\ldots,\Lambda_n)$ is a basis of $\mathfrak{g} = T_{\mathrm{id}}G$, right ? Let $m$ be such that $G \subset \mathrm{GL}_m(\mathbb{K})$ where $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ (I don't know if you work with real or complex matrices). Then $\mathfrak{g} \subset \mathrm{M}_m(\mathbb{K})$.

Then $L_g$ is a diffeomorphism of $G$ that sends $\mathrm{id}$ on $g$ thus a basis of $T_gG$ is given by the $dL_g(\mathrm{id})\Lambda_i$ and for all invertible matrices $g,h$ and all matrix $V$, $L_g(h) = gh$ hence $dL_g(h)V = gV$ so $dL_g(\mathrm{id})\Lambda_i = g\Lambda_i$. You deduce that a basis of $T_gG$ is $(g\Lambda_1,\ldots,g\Lambda_n)$ and in fact, $T_gG = g\mathfrak{g}$.

But using right multiplication, you obtain similarly that $T_gG = \mathfrak{g}g$ so $(\Lambda_1g,\ldots,\Lambda_ng)$ is an other basis of $T_gG$ and since $x \mapsto -x$ is an automorphism, $(-\Lambda_1g,\ldots,-\Lambda_ng)$ is also a basis of $T_gG$. Notice however, that we only use pushforwards, I don't see where a pullback could appear in this context.

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  • $\begingroup$ Do you happen to know a reference which talks about this as you do? $\endgroup$ Commented Jun 25, 2023 at 11:44
  • $\begingroup$ I don't, but I only use common knowledge about differential geometry, which part of the proof don't you understand ? $\endgroup$ Commented Jun 26, 2023 at 6:55

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