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This is a follow-up to this question.

For matrix Lie algebras, we can define the Lie algebra $g$ of a group $G$ as the set $T_a \in g$ that yield an element of $G$ when put into the exponential map:

$$ e^{\sum_a \alpha_a T_a} \in G $$

for some numbers $\alpha_a$

For the $SO(n)$ groups, the defining condition $O^T O=1$ means for the generators $T_a^T = -T_a$.

Now I learned in the answers to the question I linked to above that we can choose a basis for the Lie algebra, where the generators aren't antisymmetric. Why and how is this possible? If we are able to do some change of basis such that $ \tilde T_a^T \neq - \tilde T_a$ thus

$$ (e^{\sum_a \alpha_a \tilde T_a})^T e^{\sum_a \alpha_a \tilde T_a} \neq 1 $$

and therefore we do not get elements of $SO(n)$ when put into the exponential map.


EDIT: Although the comment and answer help me get closer to a satisfying understanding, I think I need a concrete example.

Therefore let's consider the Lie algebra of $SO(3)$, which consists of antisymmetric 3x3 matrices, which must have the form

$$H = \begin{pmatrix}0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0\end{pmatrix} .$$

How and why can we rewrite these basis elements such that they are no longer skew-symmetric?

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    $\begingroup$ The Lie algebra so$(n)$ as defined in your question (which I think many people would rather call the standard representation of so$(n)$) does indeed consist of only antisymmetric matrices. But this Lie algebra has other representations (algebraically isomorphic copies of it) that do not consist of only antisymmetric matrices. $\endgroup$ – Andreas Blass Jul 30 '15 at 17:15
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    $\begingroup$ @AndreasBlass Thanks for your comment. Do you know of any book or paper where such an example is discussed explicitly? What exactly do you mean by representation? In Lie theory my definition of a representation is a homomorphism to $Lin(V)$... My problem is understanding how can it be allowed to map the Lie algebra elements such that their defining feature (at least for the definition in the OP) , the skew-symmetry, gets lost and still call the elements, elements of so(n)? $\endgroup$ – JakobH Aug 1 '15 at 5:25
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Dietrich worded his answer in a way that I think is excessively confusing. What I think he means is that "diagonal" and "antisymmetric" are words that depend on a choice of basis of $\mathbb{R}^n$, not of the Lie algebra. You can get "diagonal" to mean something different by picking a basis that isn't just a differently-scaled version of the usual basis, and you can get "antisymmetric" to mean something different by picking a basis that isn't orthonormal with respect to the usual inner product. A basis is how you interpret a linear transformation $\mathbb{R}^n \to \mathbb{R}^n$ as a matrix, so if you pick a different one you get different matrices.

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  • $\begingroup$ I made an edit to the OP because I think I need a concrete example to understand this. Do you know of any easy example that shows how "antisymmetric" can "mean something different by picking a basis that isn't orthonormal with respect to the usual inner product"? $\endgroup$ – JakobH Aug 1 '15 at 5:42
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    $\begingroup$ @Jakob: rewrite all your matrices with respect to, say, the basis $e_1 = (1, 1, 1), e_2 = (0, 1, 1), e_3 = (0, 0, 1)$ of $\mathbb{R}^3$ (where the usual basis is $(1, 0, 0), (0, 1, 0), (0, 0, 1)$.) If you start with an antisymmetric matrix you won't necessarily get an antisymmetric matrix this way. $\endgroup$ – Qiaochu Yuan Aug 1 '15 at 7:05
  • $\begingroup$ @QiaochuYuan Thanks a lot! If we start with an antisym matrix acting on a vector in the standard basis $\begin{pmatrix} 0 & 0 & w \\ 0 & 0 & 0 \\ -w & 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b\\ c \end{pmatrix}$ we can switch to the basis you mentioned. Then $\begin{pmatrix} 0 & 0 & w \\ 0 & 0 & 0 \\ -w & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 &0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 &0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} a \\ b\\ c \end{pmatrix}$. $\endgroup$ – JakobH Aug 1 '15 at 7:35
  • $\begingroup$ We thus have $\begin{pmatrix} 1 & 1 &0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} a \\ b\\ c \end{pmatrix}= \begin{pmatrix} \tilde a \\ \tilde b\\ \tilde c \end{pmatrix}$, where $\tilde a,\tilde b, \tilde c$ are the coefficents w.r.t to the new basis and $\begin{pmatrix} 0 & 0 & w \\ 0 & 0 & 0 \\ -w & 0 &0 \end{pmatrix} \begin{pmatrix} 1 & 1 &0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ yields the new transformations matrix which, in general, is no longer antisymmetric. $\endgroup$ – JakobH Aug 1 '15 at 7:37

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