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Let $F$ be the free Lie algebra on $\{x,y,z\}$ and $L$ the quotient of $F$ by the ideal $I$ generated by brackets that involve at least three free generators. I have to prove that $dim(L)$ is $6$ and $L$ is nilpotent but not commutative. So if I could prove that $B:= \{\bar{x},\bar{y},\bar{z},\overline{[x,y]},\overline{[y,z]},\overline{[x,z]} \}$ is a basis of $F/I$ the statement would proved. But how can I prove that $B$ is a basis for $F/I$?

UwF
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ArthurStuart
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3 Answers3

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I would do it in two parts.

First you prove that the elements of $B$ span $F/I$ as a vector space. This is easy enough, because any longer commutator is in $I$ by assumption. So $F/I$ has dimension at most $6$.

Now to prove that they are linearly independent, you may construct a Lie algebra $L$ of dimension $6$ on three generators $X, Y, Z$ in which the elements of $I$ are relations. Such an algebra is a homomorphic image of $F/I$ by results you should have been exposed to, so $F/I$ has at least dimension $6$, so...

To construct such an $L$, consider a vector space with basis $X,Y,Z, XY, XZ, YZ$ (the last three items are just symbols), and define on it an antisymmetric binary operation by $$ [X,Y] = XY, [X, Z] = XZ, [Y, Z] = YZ, $$ and setting all $[AB, C] = 0$ (by this I mean $0 = [XY, X] = [XY, Y] = [XY, Z] = \dots$, and extending this by bilinearity.

To prove that this is a Lie algebra, you just need to show that the Jacobi identity holds. But all threefold products are trivial here, so you are done, and you have also proved in the process that this is a homomorphic image of $F/I$.

Andreas Caranti
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  • I dont' undertand the linear indipendence. Is it enought say that in the quozient the proprieties of Lie algebras (Jacoby identity and bilinearity) don't give information about a linear cobination of elements of $B$? – ArthurStuart Feb 23 '13 at 19:22
  • No, you cannot conclude lineer independence from that. – Mariano Suárez-Álvarez Feb 23 '13 at 19:25
  • @MarianoSuárez-Alvarez But I don't find a good reason to say that if $a_{1}x+a_{2}y+a_{3}z+a_{4}[x,y]+a_{5}[y,z]+a_{6}[x,z]=0$ then $a_{i}=0$... – ArthurStuart Feb 23 '13 at 19:29
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    Well, both Andreas and I expained a way to prove the linear independence of the six elements of the quotient... (In fact, that is the whole content of our answers!) – Mariano Suárez-Álvarez Feb 23 '13 at 19:30
  • @MarianoSuárez-Alvarez I'm tring to find a method that involves tensorial algebra: $F$ is the intersection of all Lie algebras contained in tensorial algebra $T(X)$ and a basis for $T(X)$ are non ordered monomials... can I use this argument? – ArthurStuart Feb 23 '13 at 19:32
  • What argument? You are trying to prove that the six elements are linearly independent in the quotient, anyways. – Mariano Suárez-Álvarez Feb 23 '13 at 19:41
  • @MarianoSuárez-Alvarez An argument that involves tensorial algebra $T(X)$ and a graduation of brackets... – ArthurStuart Feb 23 '13 at 19:44
  • The only way to tell if you can use an argument based on those things is to see the argument! And you have not explained what the argument is, only what you want to use in it. In any case, the elaboration of the idea is precisely what the theory of Shirshov-Groebner bases is. – Mariano Suárez-Álvarez Feb 23 '13 at 19:47
  • $F$ is defined as the intersection of all Lie algebras contained in tensorial algebra $T(X)$. A basis of tensorial algebra are unordered monomials. The ideal $I$ involves brackets of degree $\ge 3$ so elements of $B$ are indipendent in $F$ and so in $F/I$... Is it a correct idea? How can I ameliorate it? – ArthurStuart Feb 23 '13 at 19:52
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Consider the Lie algebra $B$ which is spanned by six linearly independent elements $x_1$, $x_2$, $x_3$, $x_{1,2}$, $x_{1,3}$ and $ x_{2,3}$, with brackets defined so that $[x_i,x_j]=x_{i,j}$ if $1\leq i<j\leq 3$ and all other brackets between elements of the basis zero. One has to check that this is a Lie algebra.

Now let $F$ be your free Lie algebra. There is a morphism of Lie algebras $\phi:F\to B$ such that $\phi(x)=x_1$, $\phi(y)=x_2$ and $\phi(z)=x_3$, and this map is surjectve because $x_1$, $x_2$, $x_3$ generate $B$. It is easy to see that your ideal $I$ is in the kernel of $\phi$, so that $\phi$ induces a Lie algebra map $\bar\phi:F/I\to B$. Since $\bar\phi$ is surjective, we see that $\dim F/I\geq6$. Moreover, since the elements of the image $X$ in $F/I$ of the set $\{x,y,z,[x,y],[x,z],[y,z]\}$ are mapped by $\bar\phi$ to linearly independent elements of $B$, the image of $X$ in $F/I$ is linearly independent.

This shows that your six elements are linearly independent. Can you show the span $F/I$?

Mariano Suárez-Álvarez
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  • If $A \in L$ then $A= \sum_{n} \sum_{a_{1}, \cdots, a_n \in {x,y,z}}$ b[a_1[a_2[...]...]. So in the quotient we'll have only sums with $x,y,z$ and their brackets with three elements. Right? – ArthurStuart Feb 18 '13 at 18:23
  • A slightly less ugly way of saying that is noting that if $V$ is the subspace of $L$ spanned by $x$, $y$ and $z$, then $L=V+[V,L]$, so that $L=V+[V,V]+[V,[V,L]]$ and $[V,[V,L]]$ is precisely your ideal. – Mariano Suárez-Álvarez Feb 18 '13 at 18:28
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A different way of doing this is to use the theory of Shirshov-Groebner bases and/or the Diamond Lemma for Lie algebras. This is explained in quite a few places; google gives me this, which is pretty nice.

Mariano Suárez-Álvarez
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