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On an infinite dimensional vector space an operator can be onto but not one-to-one (and vice versa). This arises the following question. Let $L_1$ and $L_2$ be Lie algebras (infinite dimensional, over some number field) and let $T_1:L_1\mapsto L_2$ and $T_2:L_2\mapsto L_1$ be Lie algebra homomorphisms onto. Is it true that $L_1$ and $L_2$ are isomorphic? Does it help if I know that $\ker T_1$ and $\ker T_2$ are Abelian ideals? Thank you.

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    $\begingroup$ My guess is: no. $\endgroup$ – Berci Jun 9 '16 at 22:01
  • $\begingroup$ Cross-posted to MathOF: mathoverflow.net/questions/241878 $\endgroup$ – YCor Jun 10 '16 at 18:52
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    $\begingroup$ Btw $\mapsto$ denotes the assignment. A map $X\to Y$ is denoted with a simple arrow, with the assignment $x\mapsto f(x)$ for $x\in X$. $\endgroup$ – YCor Jun 10 '16 at 19:00
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There are trivial counterexamples. For instance, let $\mathfrak{f}$ be the free Lie algebra on countably many generators (over a field $K$) and $\mathfrak{a}$ be a 1-dimensional abelian Lie algebra.

Then $\mathfrak{f}$ and $\mathfrak{f}\times\mathfrak{a}$ are both quotients of each other, but are not isomorphic (only the first one has a zero center).

Here's an example with the request that the kernels are both abelian, or even central.

Let $\mathfrak{h}$ be the (Heisenberg) Lie algebra with basis $(x,y,z)$ and only nonzero brackets $[x,y]=z$. Denote $\mathfrak{g}^\infty$ the direct sum of countably many copies of $\mathfrak{g}$.

Then $\mathfrak{h}^\infty$ and $(\mathfrak{h}\times\mathfrak{a})^\infty$ are not isomorphic (the first has center contained in the derived subalgebra and not the second). However, they are both quotient of each other with central kernel. One direction is trivial, and in the other direction, observe that $\mathfrak{a}^2$ is a quotient of $\mathfrak{h}$ with abelian kernel; hence $\mathfrak{a}^\infty$ is a quotient of $\mathfrak{h}^\infty$ with central kernel. Since $\mathfrak{h}^\infty$ is isomorphic to $\mathfrak{h}^\infty\times\mathfrak{h}^\infty$, we deduce that $\mathfrak{h}^\infty\times\mathfrak{a}^\infty$ is a quotient of $\mathfrak{h}^\infty\times\mathfrak{h}^\infty$ with central kernel.

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  • $\begingroup$ Thanks for the answer. In the example with the free Lie algebra I cannot see why $\mathfrak{f}\times\mathfrak{a}$ is a quotient of $\mathfrak{f}$. The second example is a very good one. $\endgroup$ – Bedovlat Jun 11 '16 at 9:38
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    $\begingroup$ Since $\mathfrak f\times\mathfrak a$ is countably generated it must be a quotient of the free countable lie algebra... $\endgroup$ – Myself Jun 11 '16 at 12:23
  • $\begingroup$ All right, thanks. $\endgroup$ – Bedovlat Jun 11 '16 at 21:27

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