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I am learning about Lie algebra and have come to the point where my book defines the classical Lie algebras. I am investigating $so(3)$ as I know it to the be set of $3 \times 3$ skew-symm matrices. That is, $3 \times 3$ matrices $A$ which satisfy $A^T = -A$.

My book begins by letting $S \in gl(n, \mathbb{C})$ and defining the Lie subalgebra of $gl(n, \mathbb{C})$ by $$gl_S(n,\mathbb{C}) := \{A \in gl(n,\mathbb{C}) \mid A^{t}S = -SA\}$$ Then, when $n = 2l+1$ they take $S = \begin{pmatrix}1 & 0 &0\\ 0 & 0 & I_l \\ 0 & I_l & 0 \end{pmatrix}$ and define $\mathfrak{so}(2l+1, \mathbb{C}) = gl_S(2l+1, \mathbb{C})$ and call this the orthogonal Lie algebras.

Fine. Now, I want to examine $\mathfrak{so}(3)$. This is by the above defintion $\mathfrak{so}(3) := gl_S(3, \mathbb{C})$ where $S = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1& 0 \end{pmatrix}$. Let's see what elements of this matrix look like!

Let $A = \begin{pmatrix} a_1 & a_2 &a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{pmatrix}$ and $A^t = \begin{pmatrix} a_1 & a_4 &a_7 \\ a_2 & a_5 & a_8 \\ a_3 & a_6 & a_9 \end{pmatrix}$.

Now I compute $A^t S$ and $-SA$. They are $\begin{pmatrix} a_1 & a_7 &a_4 \\ a_2 & a_8 & a_5 \\ a_3 & a_9 & a_6 \end{pmatrix}$ and $\begin{pmatrix} a_1 & a_2 &a_3 \\ a_7 & a_8 & a_9 \\ a_4 & a_5 & a_6 \end{pmatrix}$ respectively.

I have noticed that this is not exactly the original definition I've seen before. Namely, it is not $A^t = -A$. Instead, the second matrix has its second and third row swapped and the first matrix has its second and third columns swapped.

So, are these definitions equivalent? Am I allowed to swap the row/column so that is matches the $A^t = -A$ defintion?

hirotaFan
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  • @DietrichBurde Apologies, I do not understand the accepted answer there in that post. Specifically, they discuss groups of linear maps but I am not sure why. Does $\mathfrak{so}(3)$ make a group? And for the inner product, how is that defined? My book doesn't define $O(n)$ yet because I have only read about Lie algebras so far. – hirotaFan Nov 27 '19 at 21:29
  • The Lie algebras are isomorphic. The answer there uses that if the Lie groups are isomorphic so are their Lie algebras. But you can show this also directly on the level of Lie algebras. – Dietrich Burde Nov 27 '19 at 22:17

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