If $(V_{1},\pi _{1})$ and $(V_{2},\pi _{2})$ are representations of a Lie algebra $ \mathfrak {g}$, then the tensor product of these representations is the map $\pi _{1} \otimes \pi _{2}:\mathfrak {g} \to \operatorname {End} (V_{1}\otimes V_{2})$ given by
$$ (\pi _{1} \otimes \pi _{2})(X) := \pi _{1}(X) \otimes I + I \otimes \pi _{2}(X) $$
Is there a particular reason why not to define it in most naive way like for Lie groups as
$$ (\pi _{1} \otimes \pi _{2})(X) := \pi _{1}(X) \otimes \pi _{2}(X) $$
What would be in that case be violated? Or is the sole reason to to do it as in first case to make it functorial in the sense that $\pi: G \to \operatorname {GL} (V)$ induces $\pi: \mathfrak {g} \to \operatorname {End} (V)$ and in order to make the functor compatible with tensor products it should be defined as first case. Or is there a simpler reason involved why $ (\pi _{1} \otimes \pi _{2})(X) := \pi _{1}(X) \otimes \pi _{2}(X) $ is discarded?
Best Answer
Your second definition just doesn't define a Lie algebra representation at all; a Lie algebra representation must in particular be a linear map, and what you've written down isn't linear.
Conceptually $(\pi_1 \otimes \pi_2)(X) = \pi_1(X) \otimes I + I \otimes \pi_2(X)$ is an expression of the product rule. Elements $X$ of a Lie algebra correspond to infinitesimal elements of a corresponding group $\exp(\varepsilon X) = I + \varepsilon X$, and when you tensor two such infinitesimal elements together you get
$$(I + \varepsilon \pi_1(X)) \otimes (I + \varepsilon \pi_2(X)) = I + \varepsilon \left( \pi_1(X) \otimes I + I \otimes \pi_2(X) \right).$$