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In a book about compact Lie groups, I came across this notation enter image description here

and don´t know what is the meaning of the subscript $\mathbb{R}$ in the tensor product. Is it tensor product of vector spaces over $\mathbb{R}$, even if $\mathfrak{g}$ "lives" in $\mathbb{C}$?

Thank you.

Remark: $\mathfrak{g}$ = Lie algebra of some Lie subgroup of $GL(n,\mathbb{C})$.

Tereza Tizkova
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  • For example $\mathfrak{sl}2(\Bbb R)\otimes{\Bbb R} \Bbb C\cong \mathfrak{sl}_2(\Bbb C)$. – Dietrich Burde Sep 01 '21 at 11:32
  • Related: https://math.stackexchange.com/a/3975561/96384 – Torsten Schoeneberg Sep 01 '21 at 15:04
  • I'm not sure what you mean by "$\mathfrak{g}$ lives in $\mathbf{C}$". Indeed, every connected complex compact Lie group is abelian, so most likely your Lie algebra is not isomorphic to any complex Lie algebra. – YCor Sep 02 '21 at 10:35

2 Answers2

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Indeed, the subscript indicates that the tensor product is of $\mathbb R$-vector spaces (over $\mathbb C$ we have $\mathfrak g \otimes_{\mathbb C} \mathbb C \simeq \mathfrak g$.).

This makes sense because a $\mathbb C$-vector space $V$ is an $\mathbb{R}$-vector space via $t \cdot v := (t+0i)v$. In other words, we consider the restriction of scalars given by the inclusion $\mathbb R \subset \mathbb C$.

qualcuno
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  • Why is the $\mathfrak g \otimes_{\mathbb C} \mathbb C$ called "compexification" though, when it indicates tensor product of real vector spaces? – Tereza Tizkova Sep 08 '21 at 08:11
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    Notice that $\mathbb C$ is an $\mathbb R$-vector space. Thus, if $V$ is a real vector space, it makes sense to compute $\mathbb C \otimes_{\mathbb R} V$. This is now a $\mathbb{C}$-vector space, scalar multiplication is defined (by linear extension) on elementary tensors as $z \cdot a \otimes b := za \otimes b$.

    Thus we have a "canonical" procedure that, given a real vector space $V$, outputs a complex vector space associated to it. Notice that the tensor is over $\mathbb R$, not over $\mathbb C$.

     – qualcuno Sep 08 '21 at 08:26
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    (cont.) This is an instance of the general situation called "extension of scalars" whereas the situation of the original question is called "restriction of scalars". These two notions are interwined, see here. In these cases the ring map we are considering is the inclusion $\mathbb R \subset \mathbb C$. – qualcuno Sep 08 '21 at 08:29
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It means that the tensor product is considered over the ground field of real numbers. Thus, a tensor product of $\mathbb{R}$-vector spaces. This means that we must for example consider $\mathbb{C}$ as a (two-dimensional) space over $\mathbb{R}$.

J. De Ro
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