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In this book I read

Proposition A.1. The irreducible complex representations of a real Lie algebra $\mathfrak{g}$ are in one-to-one correspondence with the irreducible complex-linear representations of its complexification $\mathfrak{g}_C$

How can this be correct?

As an example take $\mathfrak{so}(3)$, which is a real Lie algebra. Naturally, the $3$-dimensional representation of $\mathfrak{so}(3)$ is a real representation

$$ \pi : \mathfrak{so}(3) \rightarrow Gl(R^3) $$

We can complexify the representation by considering

$$ \pi : \mathfrak{so}(3) \rightarrow Gl(C^3) $$

This means we have the same $3\times 3$ matrices, but now they act on complex $3$-dimensional vectors. (This is Example 5.32 at page 249 in the book I linked to above. The author writes there: "the complexification of the fundamental representation of $\mathfrak{so}(3)$ is just given by the usual $\mathfrak{so}(3)$ matrices acting on $C^3$ rather than $R^3$.

Alternatively, we can consider the complexified Lie algebra $\mathfrak{so}(3)_C$. This means we now allow complex linear combination of the $\mathfrak{so}(3)$ elements:

$$ \mathfrak{so}(3)_C= \{ x + iy | x,y \in \mathfrak{so}(3) \}$$

The representations of $\mathfrak{so}(3)_C$ are maps to complex vector spaces, for example

$$ \pi : \mathfrak{so}(3)_C \rightarrow Gl(C^3) $$

These are the complex linear combinations of the usual $\mathfrak{so}(3)$ matrices acting on complex vectors.

How can this representation be in "in one-to-one correspondence" to $ \pi : \mathfrak{so}(3) \rightarrow gl(C^3) $?

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    $\begingroup$ Strictly speaking, the author should be talking about the sets of isomorphism classes of irreducible complex representations of $\mathfrak{g}$ vs. its complexification. There is a natural function from one set to the other given by complexifying the action of $\mathfrak{g}$, and the claim is that this is a bijection. This function sends the first complex representation you wrote down to the second one. There's no problem here. $\endgroup$ Commented Aug 25, 2015 at 9:09

1 Answer 1

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An irreducible representation $\phi:\mathfrak{g}_\mathbb{C} \rightarrow GL(\mathbb{C,n})$ is a Lie algebra homomorphism by definition. That is, it's linear.

So, to consider your example, we see that for $x,y \in \mathfrak{so}(3)$,

\begin{align} \pi(x+iy) = \pi(x)+i\pi(y) \end{align}

which shows that working with

\begin{align} \pi: \mathfrak{so}(3) \rightarrow GL(\mathbb{C,3}) \end{align}

is the same as working with

\begin{align} \pi: \mathfrak{so}(3)_\mathbb{C} \rightarrow GL(\mathbb{C,3}) \end{align}

(if this doesn't convince you, try to construct a counter example to the proposition and you'll see that it's impossible precisely because any irreducible representation is a homomorphism).

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    $\begingroup$ where are you using the irreducibility of the representations? $\endgroup$
    – glS
    Commented Oct 14, 2020 at 10:07
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    $\begingroup$ The representation doesn't need to be irreducible. The question was asked with irreducible representations in mind, and if irreducible representations are linear, so are reducible representations. $\endgroup$ Commented Jun 19, 2022 at 9:44

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