MHB What is the Definition of a Derivation for a Lie Algebra?

[topsquark]

Loosely speaking a derivation D is defined as a function on an algebra A that has the property D(ab) = (Da)b + a(Db).

Now, if we define the map [math]ad_x: y \mapsto [x,y] [/math] and apply this to the Jacobi identity we get [math]ad_x[y,z] = [ ad_x(y),z ] + [ y, ad_x(z) ] [/math]. This does not look quite like the definition of the derivation given above. It is considered a derivation because of the ordering inside the brackets? Or does is this simply the definition of a derivation for a Lie algebra?

Ooh! Wait a minute. The brackets are there because multiplication in the Lie algebra is given by [math] [,]: L \times L \mapsto L[/math]? (My notes are not clear that this is to represent multiplication.)

-Dan

 

[Euge]

Loosely speaking a derivation D is defined as a function on an algebra A that has the property D(ab) = (Da)b + a(Db).

Now, if we define the map [math]ad_x: y \mapsto [x,y] [/math] and apply this to the Jacobi identity we get [math]ad_x[y,z] = [ ad_x(y),z ] + [ y, ad_x(z) ] [/math]. This does not look quite like the definition of the derivation given above. It is considered a derivation because of the ordering inside the brackets? Or does is this simply the definition of a derivation for a Lie algebra?

Ooh! Wait a minute. The brackets are there because multiplication in the Lie algebra is given by [math] [,]: L \times L \mapsto L[/math]? (My notes are not clear that this is to represent multiplication.)

-Dan

Hi Dan,

It is common to use brackets to represent multiplication in a Lie algebra. The identity you have above for $ad_x$ shows that $ad_x$ is a derivation.

Here's something to keep in mind. Suppose $A$ is an associative algebra. Define $[x, y]$ to be the additive commutator $xy - yx$. Then $[x, x] = 0$ for all $x\in A$ and Jacobi's identity is satisfied with $[\, ,\,]$. So $[\, ,\,]$ is a Lie bracket on $A$, and $(A, [\, ,\,])$ is a Lie algebra. This gives a connection between associative algebras and Lie algebras.

 

[topsquark]

Thank you for the reply. I didn't think of it until later but this explains the notations in some of the more arcane set of proofs in my QFT texts.

One further question. The definition of a Lie group on set L gives a multiplicative operation [math]m: L \times L \to L[/math]. Does this imply a unique form of [,] in the Lie algebra or can we have more than one Lie algebra corresponding to a given Lie group?

-Dan

Edit:

This gives a connection between associative algebras and Lie algebras.

There are non-associative algebras??

 

[Euge]

Thank you for the reply. I didn't think of it until later but this explains the notations in some of the more arcane set of proofs in my QFT texts.

One further question. The definition of a Lie group on set L gives a multiplicative operation [math]m: L \times L \to L[/math].

What do you mean by a "Lie group on set L"?

There are non-associative algebras??

Yes. Here's one that you'll be familiar with. Consider $\Bbb R^3$ with multiplication defined by the cross product. This defines an algebra over $\Bbb R$. Let $\textbf{u} = (2, 3, 1)$, $\textbf{v} = (1, 0, 2)$, and $\textbf{w} = (0, 0, 1)$. Then $\textbf{u} \times (\textbf{v} \times \textbf{w}) = (2, 3, 1) \times (0, -1, 0) = (1, 0, -2)$ and $(\textbf{u} \times \textbf{v}) \times \textbf{w} = (6,-3,-3) \times (0,0,1) = (-3,-6,0)$. Therefore, $\textbf{u} \times (\textbf{v} \times \textbf{w}) \neq (\textbf{u} \times \textbf{v}) \times \textbf{w}$. Geometrically speaking, the cross products $\textbf{u} \times (\textbf{v} \times \textbf{w})$ and $(\textbf{u} \times \textbf{v}) \times \textbf{w}$ lie in different planes in general, so they are usually not equal.

To every algebra $A$ there corresponds a (multilinear) ternary operation $[\cdot, \cdot, \cdot] : A \times A \times A \to A$, called the $\textit{associator}$, defined by $[x,y,z] = (xy)z - x(yz)$. Of course, the associator of $A$ is trivial if and only if $A$ is associative. Make sure not to assume algebras are associative in your texts unless mentioned otherwise.

 

[topsquark]

What do you mean by a "Lie group on set L"?

Sorry. It was a badly worded question anyway but apparently I was so set on assuming that a Lie algebra was created from a Lie group that I didn't see that the definition says that a Lie algebra is created from a vector space. Please ignore the question.

-Dan