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Lie algebras are usually classified modulo isomorphisms of the underlying vector space, i.e. invertible changes of basis, which are $GL(N)$ maps of the generators $X'^i = M^i{}_j \, X^j$. For example, there are only two classes of 2D Lie algebras: the Abelian one $[X,Y]=0$ and the non-Abelian one $[X,Y]=Y$. A generic 2D Lie algebra would be $[X,Y]=a \, X+b \, Y$, but unless $a=b=0$, there always is a $GL(2)$ transformation like above, such that $[X',Y']=Y'$. Notice that such a transformation in general is not an algebra homomorphism: it does not preserve the Lie brackets (i.e. it does not leave the structure constants unchanged).

Lie algebra automorphisms are homomorphisms from the algebra to itself, that is, linear maps $X^i \to \varphi (X^i)$ such that $[\varphi(X^i),\varphi(X^j)]=\varphi([X^i,X^j])$. They are a subset of the changes of basis that preserve the Lie brackets, or, equivalently, that preserve the structure constants. In the example above, an automorphism of the Lie algebra $[X,Y]=Y$ is the map $X'= X$, $Y' = \alpha \, Y$ with $\alpha \neq 0$. This map is such that $[X',Y']=Y'$. Clearly not all $GL(N)$ changes of basis are automorphisms, and they form a nontrivial subgroup of $GL(N)$.

Now, when Lie bialgebras are studied, their classification is usually done up to automorphisms of the underlying Lie algebra. That is, if I want to classify all Lie bialgebras on the Lie algebra $\mathfrak g$, two of them are considered equivalent if there is an automorphism of $\mathfrak g$ that sends the cobracket of one into the other. My question is, why not consider equivalent two Lie bialgebras on the same vector space if they are related by a generic $GL(N)$ change of basis? Why limit it to automorphisms? The way that $GL(N)$ changes of basis act on the structure constants of the algebra and of the coalgebra, they all preserve the validity of the Jacobi, co-Jacobi and compatibility rules, whether they are automorphisms or not. In other words, any change of basis sends Lie bialgebras into Lie bialgebras, so it seems natural to me to consider equivalent two Lie bialgebras that are related by any $GL(N)$ transformation. What am I missing?

Spinoro
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    But there is a Lie algebra homomorphism (even an isomorphism) between the algebras with $[X,Y]=aX+bY$ and the algebra with $[x,y]=y$ for $(a,b)\neq (0,0)$. Such an $f$ satisfies $f([A,B])=[f(A),f(B)]$ for all $A,B$. – Dietrich Burde Jan 08 '25 at 17:21
  • Yeah, the linear map you are describing isn't a linear map,at all. If $X,Y$ is a basis with $[X,Y]=aX+bY,$ then there exists another basis $x',Y'$ with $[X',Y']=Y'.$ Sure, you can define a linear map $X\mapsto X', Y\mapsto Y',$ but that has nothing to do with the Lie algebra, and is not a homomorphism. What you are noticing is that the two algebras, along with specific bases, are not isomorphic. But ${X,Y\mid [X,Y]=aX+bY}$ is isomorphic to ${X,Y\mid [X,Y]=Y}z.$ The isomorphism is just not $X\mapsto X, Y\mapsto Y.$ – Thomas Andrews Jan 08 '25 at 17:39
  • @DietrichBurde Ah wait, I must have been confused as to what constitutes a Lie algebra homomorphism... for example, in your example, if $[X,Y] = a X+bY$ and $[x,y]=y$, I can find the map: $\varphi(X) =b x + y$, $\varphi(Y) = - a x$, and then $[\varphi(X) , \varphi(Y)] = a y = a \varphi(X) + b \varphi(Y) = \varphi(a X+bY) = \varphi([X,Y])$. So this is a map from the algebra $(X,Y)$ to the algebra $(x,y)$, that is an homomorphism. But now, are you saying that, in order to classify the 2D Lie algebras, I need to only stick to linear transformations that are homomorphisms? – Spinoro Jan 08 '25 at 18:07
  • Yes, a Lie algebra isomorphism is just a bijective linear map $f$ satisfying $f([x,y])=[f(x),f(y)]$. Therefore we have only one non-abelian Lie algebra in dimension $2$, up to isomorphism. – Dietrich Burde Jan 08 '25 at 19:12
  • Ok, thanks. But then is any $GL(N)$ change of basis a Lie algebra isomorphism? – Spinoro Jan 08 '25 at 20:34
  • My main doubt here is: when people classify the Lie bialgebra structures on a given Lie algebra $\mathfrak{g}$, they usually do it up to automorphisms of $\mathfrak{g}$. This, in my understanding, means up to linear transformations of the basis that leave the Lie brackets unchanged. But what if I make a general change of basis? I get a new Lie bialgebra. Wouldn't you say that this new Lie bialgebra is equivalent to the original one? – Spinoro Jan 09 '25 at 18:10

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OK, I figured it out. A general change of basis acts on both the Lie bracket and the cocommutator (if you want, it transforms both the structure constants of the algebra and the coalgebra), and gives rise to a Lie bialgebra homomorphism (isomorphism if it's invertible). So, we consider Lie bialgebras that are related by such a transformation as equivalent. However, when classifying all the Lie bialgebra structure over a particular Lie algebra, we can keep the Lie brackets fixed (i.e. choose a particular basis for the Lie algebra and stick with it), and consider all possible structure constant for the coalgebra. This way, two Lie bialgebras with same structure constants for the algebra but different structure constants for the coalgebra are ensured to be inequivalent, unless there is a change of basis between the two coalgebras, that is also an automorphism of the Lie algebra. In that case, the same change of basis that relates the two coalgebras is also a change of basis for the whole Lie bialgebra. For this reason, when classifying Lie bialgebra structures over a given Lie algebra, we need to quotient by the automorphism group of the Lie algebra.

Spinoro
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