📂Representation Theory

Direct Sum of Lie Algebras

Definition¹

Let $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ be Lie algebras. The direct sum $\mathfrak{g}$ of $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ is defined as the direct sum as vector spaces.

$$ \mathfrak{g} = \mathfrak{g}_{1} \oplus \mathfrak{g}_{2} $$

The bracket of $\mathfrak{g}$ is as follows.

$$ \left[ (X_{1}, X_{2}), (Y_{1}, Y_{2}) \right]_{\mathfrak{g}} = \left([X_{1}, Y_{1}]_{\mathfrak{g}_{1}}, [X_{2}, Y_{2}]_{\mathfrak{g}_{2}}\right) \tag{1} $$

Let $\mathfrak{g} = \mathfrak{g}_{1} \oplus \mathfrak{g}_{2}$ be a Lie algebra, and let $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$ be Lie subalgebras of $\mathfrak{g}$. If the following equation holds, then $\mathfrak{g}$ is said to decompose as the Lie algebra direct sum of $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$.

$$ [X_{1}, X_{2}] = 0 \quad \text{for all } X_{1} \in \mathfrak{g}_{1}, X_{2} \in \mathfrak{g}_{2} $$

Explanation

The bracket defined by $(1)$ means that it acts independently, without any cross terms between $\mathfrak{g}_{1}$ and $\mathfrak{g}_{2}$. Moreover, it makes $\mathfrak{g}_{1} \oplus \mathfrak{g}_{2}$ a Lie algebra.

• Bilinear

From the bilinearity of the brackets of $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$, the bracket of the direct sum is also bilinear. Since in the direct sum of vector spaces $a(X_{1}, X_{2}) + b(Y_{1}, Y_{2}) = (aX_{1} + bY_{1}, aX_{2} + bY_{2})$, $$ \begin{align*} & \Big[ a(X_{1}, X_{2}) + b(Y_{1}, Y_{2}), (Z_{1}, Z_{2}) \Big]_{\mathfrak{g}} \\ &= \Big[ (aX_{1} + bY_{1}, aX_{2} + bY_{2}), (Z_{1}, Z_{2}) \Big]_{\mathfrak{g}} \\ &= \left( \left[ aX_{1}+bY_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, \left[ aX_{2}+bY_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right) \\ &= \left( a\left[ X_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}+b\left[ Y_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, a\left[ X_{2}, Z_{2} \right]_{\mathfrak{g}_{2}}+b\left[ Y_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right) \\ &= \left( a\left[ X_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, a\left[ X_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right) + \left( b\left[ Y_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, b\left[ Y_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right)\\ &= a\left( \left[ X_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, \left[ X_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right) + b\left( \left[ Y_{1}, Z_{1} \right]_{\mathfrak{g}_{1}}, \left[ Y_{2}, Z_{2} \right]_{\mathfrak{g}_{2}} \right)\\ &= a \left[ (X_{1}, X_{2}), (Z_{1}, Z_{2}) \right]_{\mathfrak{g}} + b \left[ (Y_{1}, Y_{2}), (Z_{1}, Z_{2}) \right]_{\mathfrak{g}}\\ \end{align*} $$

• Antisymmetry

From the antisymmetry of the brackets of $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$, the bracket of the direct sum acquires antisymmetry. $$ \begin{align*} \left[ (X_{1}, X_{2}), (Y_{1}, Y_{2}) \right]_{\mathfrak{g}} &= \big([X_{1}, Y_{1}]_{\mathfrak{g}_{1}}, [X_{2}, Y_{2}]_{\mathfrak{g}_{2}}\big) \\ &= \big(-[Y_{1}, X_{1}]_{\mathfrak{g}_{1}}, -[Y_{2}, X_{2}]_{\mathfrak{g}_{2}}\big) \\ &= -\big([Y_{1}, X_{1}]_{\mathfrak{g}_{1}}, [Y_{2}, X_{2}]_{\mathfrak{g}_{2}}\big) \\ &= -\left[ (Y_{1}, Y_{2}), (X_{1}, X_{2}) \right]_{\mathfrak{g}} \\ \end{align*} $$

• Jacobi identity

Since the brackets of $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ satisfy the Jacobi identity, the bracket of $\mathfrak{g}$ satisfies it as well. Computing one term first gives the following.

$$ \begin{align*} \Big[ \left[ (X_{1}, X_{2}), (Y_{1}, Y_{2}) \right]_{\mathfrak{g}}, (Z_{1}, Z_{2}) \Big]_{\mathfrak{g}} &= \Big[ \big([X_{1}, Y_{1}]_{\mathfrak{g}_{1}}, [X_{2}, Y_{2}]_{\mathfrak{g}_{2}}\big), (Z_{1}, Z_{2}) \Big]_{\mathfrak{g}} \\ &= \Big( \big[ [X_{1}, Y_{1}]_{\mathfrak{g}_{1}}, Z_{1} \big]_{\mathfrak{g}_{1}}, \big[ [X_{2}, Y_{2}]_{\mathfrak{g}_{2}}, Z_{2} \big]_{\mathfrak{g}_{2}} \Big) \end{align*} $$

Computing the remaining two terms in the same manner and adding all three terms, each component becomes the Jacobi identity in $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$ and hence equals $0$.

$$ \begin{align*} &\Big[ \left[ (X_{1}, X_{2}), (Y_{1}, Y_{2}) \right]_{\mathfrak{g}}, (Z_{1}, Z_{2}) \Big]_{\mathfrak{g}} + \Big[ \left[ (Y_{1}, Y_{2}), (Z_{1}, Z_{2}) \right]_{\mathfrak{g}}, (X_{1}, X_{2}) \Big]_{\mathfrak{g}} + \Big[ \left[ (Z_{1}, Z_{2}), (X_{1}, X_{2}) \right]_{\mathfrak{g}}, (Y_{1}, Y_{2}) \Big]_{\mathfrak{g}} \\ &= \left( \begin{aligned} &\big[ [X_{1}, Y_{1}]_{\mathfrak{g}_{1}}, Z_{1} \big]_{\mathfrak{g}_{1}} + \big[ [Y_{1}, Z_{1}]_{\mathfrak{g}_{1}}, X_{1} \big]_{\mathfrak{g}_{1}} + \big[ [Z_{1}, X_{1}]_{\mathfrak{g}_{1}}, Y_{1} \big]_{\mathfrak{g}_{1}}, \\ &\big[ [X_{2}, Y_{2}]_{\mathfrak{g}_{2}}, Z_{2} \big]_{\mathfrak{g}_{2}} + \big[ [Y_{2}, Z_{2}]_{\mathfrak{g}_{2}}, X_{2} \big]_{\mathfrak{g}_{2}} + \big[ [Z_{2}, X_{2}]_{\mathfrak{g}_{2}}, Y_{2} \big]_{\mathfrak{g}_{2}} \end{aligned} \right) \\ &= (0, 0) \end{align*} $$