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This bit of text comes from Lee's Introduction to Smooth Manifolds.

I don't see why (8.15) holds. Note first of all that Lee assumes the Einstein summation convention, while I will not in my formulation. I would think that we have $$ A^L\vert_X=\sum_{i=1}^n\sum_{j=1}^n X^i_j A^i_j\frac\partial{\partial X^i_j}\bigg\vert_X, $$ instead of $$ A^L\vert_X=\sum_{k=1}^n\sum_{i=1}^n\sum_{j=1}^n X^i_j A^j_k\frac\partial{\partial X^i_k}\bigg\vert_X, $$ since I think it holds that $$ d(L_X)_{I_n}\left(\frac\partial{\partial X^i_j}\bigg\vert_{I_n}\right)=X^i_j\frac\partial{\partial X^i_j}\bigg\vert_X. $$ I argued this using the coordinate reprsentation of the differential, which is given for an arbitrary smooth map $F\colon M\to N$ by $$ dF_p\left(\frac\partial{\partial x^i}\bigg\vert_p\right)=\frac{\partial\hat F^i}{\partial x^j}\bigg\vert_{\hat p}\frac\partial{\partial y^j}\bigg\vert_{F(p)}, $$ where $(x^i)$ are local coordinates for some open $U\ni p$, and $(y^i)$ are local coordinates for some open $V\ni F(p)$.

Hence, if we take $(E^i_j)$ as our basis for $\operatorname M_n(\mathbb R)$, then $E^i_j$ is mapped by $L_X$ to $X^j_i$. And therefore $$ \frac{\partial(L_X)^i_j}{\partial x^k_l}=\delta_{ik}\delta_{jl} X^j_i. $$ Note that also here, I don't assume Einstein summation convention.

So I don't see why (8.15) holds... could someone clarify?

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  • $\begingroup$ You can only sum over an index once when you use the Einstein summation convention. You have to be willing to change indices of summation when you work with this stuff. Don't use $i$ and $j$ more than twice each. $\endgroup$ May 4, 2020 at 21:55
  • $\begingroup$ @TedShifrin In the formulas that I give (not the one from Lee), I am not using Einstein summation convention. $\endgroup$
    – Sha Vuklia
    May 4, 2020 at 21:56
  • $\begingroup$ Well, I have given the correct answer below, but I invite you to make sense of your $A^L|_X$ formula. Moreover, your displayed formula makes no sense because we have no idea what $A^i_j$ is. All we know about is the matrix $X$. $\endgroup$ May 4, 2020 at 21:59
  • $\begingroup$ I still invite you to make sense of your formula. If you're not using Einstein summation convention, please write down (with summation symbols) exactly what this means. $\endgroup$ May 4, 2020 at 22:03
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    $\begingroup$ My correction was a little hasty. If $X = \sum_{k,l} X^k_l E^l_k $ then $X E^i_j = \sum_l X^i_l E^l_j$. That's actually what Prof. Shifrin wrote below. Anyway perhaps you made an error keeping track of your indices? $\endgroup$
    – Max
    May 4, 2020 at 22:55

2 Answers 2

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Note that when you left multiply by $X$, you now have the matrix whose $ik$-entry is $X^i_j A^j_k$. If we want its $ij$-entry, we should change letters around and write $X^i_\ell A^\ell_j$. This then becomes the coefficient of $\partial/\partial X^i_j$.

Bottom line, the matrix $A$ at the identity element left-translates to the matrix $XA$ at the point $X$. Writing that in terms of the standard basis is precisely what Lee has done.

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  • $\begingroup$ Why would we include $A$ when determining what the $ik$-entry of $d(L_X)$ is? I thought that since $L_X$ is a linear operator, we just have to determine what $L_X$ does with $E_{ik}$ to determine the $ik$-entry, and that is in this case $X_{ik}$, since $L_X(E_{ik})=XE_{ik}=X_{ik}$. $\endgroup$
    – Sha Vuklia
    May 4, 2020 at 22:05
  • $\begingroup$ Because we're finding the left-invariant vector field whose value at the identity is $A$!! $\endgroup$ May 4, 2020 at 22:07
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Maybe you got confused as me when i first read it. So maybe it's better to denote the point $X$ in $L_X$ as a more suggestive notation $L_{X_0}$, which means that $X_0$ is fix.

Suppose that $(X^i_j)$ denote the standard global coordinate for $\text{GL}(n, \mathbb{R})$ and let $X_0=(X_0)^i_j \in \text{GL}(n, \mathbb{R})$ be a chosen point where we want to compute $A^\text{L}|_{X_0}$. The matrix representation of the map $L_{X_0} : \text{GL}(n, \mathbb{R}) \to \text{GL}(n, \mathbb{R})$, $X \mapsto X_0X$ in this coordinates is $$ (X^i_j) \mapsto (L_{X_0}X)^i_j = (X_0)^i_k X^k_j. $$ So \begin{align*} A^{L}|_{X_0} &= d(L_{X_0})_{I_n}\bigg( A^i_j \frac{\partial}{\partial X^i_j}\bigg|_{I_n} \bigg) = A^i_j \, \bigg(\frac{\partial (L_{X_0})^k_m}{\partial X^i_j}\bigg)_{I_n} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0} \\ &= A^i_j \, \bigg(\frac{\partial (X_0X)^k_m}{\partial X^i_j}\bigg)_{I_n} \frac{\partial}{\partial X^k_m}\bigg|_{X_0} \\ &= A^i_j \, \bigg(\frac{\partial ((X_0)^k_l X^l_m)}{\partial X^i_j}\bigg)_{I_n} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0} \\&= A^i_j (X_0)^k_l \delta_{il} \delta_{jm} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0}\\ &= (X_0)^i_j A^j_k \frac{\partial}{\partial X^i_k}\bigg|_{X_0}. \end{align*}

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  • $\begingroup$ Ah, I see now that the differential of $L_{X_0}$ at the identity is given by your expression, but I have to figure out what mistake I was making when working with $(E^i_j)$. Normally, when we consider $\frac{\partial F^i}{\partial x^j}\big\vert_x$, that is just the $ij$-th entry of $DF(x)$. So my reasoning was to determine $D(L_{X_0})_{I_n}$. Since $L_{X_0}$ is linear, this is the same as determining the matrix represenation of $L_{X_0}$ (in the $(E^i_j)$ basis). So I reckoned, $E^i_j$ is send by $L_{X_0}$ to $(X_0)^j_i$. But this way I don't get the summation from matrix multiplication, hm.. $\endgroup$
    – Sha Vuklia
    May 4, 2020 at 22:51
  • $\begingroup$ @ShaVuklia What is $E^i_j$? $\endgroup$ May 4, 2020 at 22:53
  • $\begingroup$ The matrix with $1$ on the $ij$-th entry, and zeros elsewhere. It's what I believe to be the standard basis on $\operatorname M_n(\mathbb R)$. $\endgroup$
    – Sha Vuklia
    May 4, 2020 at 22:54
  • $\begingroup$ Ohhhh, never mind. Max helped me out. I made a stupid error, thinking $XE^i_j=X^j_i$. $\endgroup$
    – Sha Vuklia
    May 4, 2020 at 22:56
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    $\begingroup$ Ok. So you mean $D(L_{X_0})_{I_n}(E^i_j) = D(L_{X_0})_{I_n}(\partial/\partial X^i_j)$ by standard identification. $\endgroup$ May 4, 2020 at 22:58

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