I would appreciate help showing the complexification of these Lie algebras are isomorphic:

$$\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}\cong \mathfrak{u}(n)_{\mathbb{C}}$$

where $\mathfrak{gl}(n,\mathbb{R})= M_n(\mathbb{R})$ and $\mathfrak{u}(n)=\lbrace X\in M(\mathbb{C}):X+X^{*}=0\rbrace$

Revealing my ignorance, I especially am confused regarding the complexification of $\mathfrak{u}(n)$ since it already is comprised of complex matrices. And how can $\mathfrak{gl}(n,\mathbb{R})_{\mathbb{C}}$ be isomorphic to $\mathfrak{u}(n)_{\mathbb{C}}$ in view of the stipulation regarding elements of $\mathfrak{u}(n) $that $X+X^{*}=0$



Let $H_i$ denote the $n$-by-$n$ diagonal matrix with $1$ at the $(i,i)$-entry and $0$ elsewhere. For $i,j=1,2,\ldots,n$ with $i<j$, let $X_{i,j}$ be the matrix with $1$ at the $(i,j)$-entry and $0$ elsewhere, whilst $Y_{i,j}$ denotes $X_{i,j}^\top$. Note that the matrices $H_i$'s, $X_{i,j}$'s, and $Y_{i,j}$'s form a Chevalley basis of the Lie algebra $\mathfrak{gl}(n,\mathbb{C})$, noting that $$\mathfrak{gl}(n,\mathbb{R})_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})\,.$$ (An isomorphism $\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\mathfrak{gl}(n,\mathbb{R})\cong\mathfrak{gl}(n,\mathbb{C})$ should be easy to find).

A basis for $\mathfrak{u}(n)\subseteq\mathfrak{gl}(n,\mathbb{C})$ is given by $\tilde{h}_i$, $\tilde{x}_{i,j}$, and $\tilde{y}_{i,j}$ where $i,j=1,2,\ldots,n$ and $i<j$, where $$\tilde{h}_i:=\sqrt{-1}\,H_i\,,\,\,\tilde{x}_{i,j}:=\frac{1}{\sqrt{2}}\,\left(X_{i,j}-Y_{i,j}\right)\,,\text{ and }\tilde{y}_{i,j}:=\frac{\sqrt{-1}}{\sqrt{2}}\,\left(X_{i,j}+Y_{i,j}\right)\,.$$ Now, set $$h_i:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{h}_i\,,\,\,x_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{x}_{i,j}\,,\text{ and }y_{i,j}:=1\,\underset{\mathbb{R}}{\otimes}\,\tilde{y}_{i,j}\,.$$ Then, we see that the elements $h_i$'s, $x_{i,j}$'s, and $y_{i,j}$'s form a Chevalley basis of $\mathfrak{u}(n)_\mathbb{C}=\mathbb{C}\,\underset{\mathbb{R}}{\otimes}\,\mathfrak{u}(n)$. Consequently, an isomorphism $\varphi:\mathfrak{gl}(n,\mathbb{C})\to\mathfrak{u}(n)_\mathbb{C}$ is the linear extension of the assignments $$\varphi\left(H_i\right)=h_i\,,\,\,\varphi\left(X_{i,j}\right)=x_{i,j}\,,\text{ and }\varphi\left(Y_{i,j}\right)=y_{i,j}$$ for $i,j=1,2,\ldots,n$ with $i<j$.

P.S.: I said something wrong. I called the bases I provided Chevalley bases, but they are actually not.

  • $\begingroup$ Thanks for your answer. Unfortunately it is somewhat over my head - especially the tensors. If I may say, the isomorphism was in a set of notes primarily for undergraduates in a Lie theory course. It was mentioned as "an easy exercise." Subsequently. $\mathfrak{u}(n)_{\mathbb{C}}\cong \mathfrak{gl}(n,\mathbb{C})$ was proved along the lines as you did with $\tilde{x}_{i,j}:=\frac{1}{\sqrt{2}}\,\left(X_{i,j}-Y_{i,j}\right)\,,\text{ and }\tilde{y}_{i,j}:=\frac{\sqrt{-1}}{\sqrt{2}}\,\left(X_{i,j}+Y_{i,j}\right)\,.$ $\endgroup$ – user12802 May 20 '17 at 23:31

This answer is incorrect - See comment below

In an attempt to answer my own question, can I not say that two finite dimensional vector spaces with the same dimension are isomorphic?

Thus $\dim \mathfrak{gl}(n,\mathbb{R})= \dim \mathfrak{u}(n)= n^2$ And their respective complexifications have dimension $2n^2$.

  • 1
    $\begingroup$ No, you can't. Isomorphisms of vector spaces are usually not isomorphisms of Lie algebras. So, you only know that the two Lie algebras are isomorphic as vector spaces, but that says nothing about their Lie algebra structures. For example, the Heisenberg Lie algebra $\mathfrak{H}(\mathbb{C})$ over $\mathbb{C}$ is $3$-dimensional, just as $\mathfrak{sl}(2,\mathbb{C})$. But they are not isomorphic, as $\mathfrak{H}(\mathbb{C})$ is nilpotent and $\mathfrak{sl}(2,\mathbb{C})$ is simple. $\endgroup$ – Batominovski May 23 '17 at 0:17

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