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Assume that $\mathcal{A}$ is a Lie $\mathbb{C}$-algebra. Also denote the Lie $\mathbb{C}$-algebra $\mathcal{D}=\mathbb{C}[x_1,\ldots, x_n]\partial_{x_1}\oplus\ldots\oplus\mathbb{C}[x_1,\ldots, x_n]\partial_{x_n}$. My questions are as follows

Let $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$, $\mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}(t)$ are isomorphic $\mathbb{C}(t)$-Lie algebras.

  1. Is it true that $\mathcal{A}, \mathcal{D}$ are isomorphic $\mathbb{C}$-Lie algebras?

  2. Is it true that $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}[t], \mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}[t]$ are isomorphic $\mathbb{C}[t]$-Lie algebras?

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  • $\begingroup$ It's not tensor product of Lie algebras, it's extension of scalars of Lie algebras. Also I assume that the isomorphism in the assumption is an isomorphism of Lie algebras over $C(t)$? $\endgroup$
    – YCor
    Commented Jul 23, 2020 at 7:35
  • $\begingroup$ Yes, you are right $\endgroup$
    – solver6
    Commented Jul 23, 2020 at 7:36

1 Answer 1

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Put $\mathcal{A}'=\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and similarly for $\mathcal{D}'$. Choose an isomorphism $f\colon\mathcal{D}'\to\mathcal{A'}$.
Choose a (countable) basis $\mathcal{D}_0=\{d_i:i\in\mathbb{N}\}$ for $\mathcal{D}$ over $\mathbb{C}$. Then $f(\mathcal{D}_0)$ is a basis for $\mathcal{A}'$ over $\mathbb{C}(t)$ but $\dim_{\mathbb{C}}(\mathcal{A}')=\dim_{\mathbb{C}}(\mathcal{A})$ so we can also choose a countable basis $\mathcal{A}_0=\{a_i:i\in\mathbb{N}\}$ for $\mathcal{A}$ over $\mathcal{C}$. We must have $[d_i,d_j]=\sum_kp_{ijk}d_k$ and $[a_i,a_j]=\sum_kq_{ijk}a_k$ for some structure constants $p_{ijk},q_{ijk}\in\mathbb{C}$. Now let $K$ be a countable subfield of $\mathbb{C}$ containing all these structure constants, and also all the constants needed to ensure that $f(\mathcal{A}_0)\subseteq K(t).\mathcal{D}_0$ and $f^{-1}(\mathcal{D}_0)\subseteq K(t).\mathcal{A}_0$. Put $\mathcal{A}_1=K.\mathcal{A}_0$ and $\mathcal{D}_1=K.\mathcal{D}_0$, so these are Lie algebras over $K$ that become isomorphic over $K(t)$. If $\mathbb{C}$ were algebraic over $K$ then it would be countable, which is false. Thus, we can choose an embedding $i\colon K(t)\to\mathbb{C}$ extending the identity on $K$. By applying this to the coefficients of $f$, we obtain an isomorphism $\mathcal{D}\simeq\mathcal{A}$.

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  • $\begingroup$ Thank you. I don't have enough knowledge about structure constants, so right now I cannot check your proof completely. From the answer we have that 1), 2) are true, but what about 3)? $\endgroup$
    – solver6
    Commented Jul 23, 2020 at 9:52
  • $\begingroup$ If by structure constants you mean coordinates of commutators of elements from a countable basis then it looks as a correct proof for me. I need to think more about it to mark as accepted $\endgroup$
    – solver6
    Commented Jul 23, 2020 at 10:00
  • $\begingroup$ So you proved the more general fact that if $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ is isomorphic as $\mathbb{C}(t)$-Lie algebra to $\mathcal{B}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and $\dim_{\mathbb{C}}\mathcal{A}=\dim_{\mathbb{C}}\mathcal{B}$-countable then $\mathcal{A}$ is isomorphic to $\mathcal{B}$ as $\mathbb{C}$-Lie algebra. Is there any known reference to this fact existed in literature? $\endgroup$
    – solver6
    Commented Jul 23, 2020 at 10:09
  • $\begingroup$ I wonder why a similar argument would not work with $\mathbb C((t))$... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 23, 2020 at 10:38
  • $\begingroup$ @მამუკაჯიბლაძე the $t$-adic topology on $K((t))$ doesn't play well with embedding into $\mathbb C$. $\endgroup$
    – Joshua Mundinger
    Commented Jul 23, 2020 at 16:49

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