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It is well-known that the algebra $ so(4,1) $ admits the Killing splitting as vector spaces as follows: $ so(4,1)=so(3,1) \oplus \mathbb{R}^{3,1} $. In the same context, is there a Killing splitting of the algebra $ gl(5,\mathbb{R}) $ of the group $ GL(5,\mathbb{R}) $?

Oussama Abdelghafour
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    By Killing splitting do you mean a splitting into two orthogonal subspaces under the Killing form? In which case, basically, yes. The important thing to note here though is that the Killing form is no longer nondegenerate here. The span of the identity is orthogonal to everything including itself. Note there are of course many different ways to split it (as indeed there are for $\mathfrak{so(4,1)}$) – Callum Dec 08 '23 at 13:16
  • Yes, I do. Can you give me just one example, please? – Oussama Abdelghafour Dec 08 '23 at 19:08
  • Sure here's a whole family: let $\mathbb{R}^n = V_1 \oplus \cdots \oplus V_k$ be a decomposition. Then take $\mathfrak{h} = \mathfrak{gl}(V_1) \oplus \cdots \oplus \mathfrak{gl}(V_k) $ and $\mathfrak{m} = \bigoplus_{i\neq j} \mathrm{Hom}(V_i, V_j)$ then $\mathfrak{gl}(\mathbb{R}^n) =\mathfrak{h}\oplus\mathfrak{m}$. Otherwise said $\mathfrak{h}$ is the block diagonal part and $\mathfrak{m}$ is the block off-diagonal part. – Callum Dec 08 '23 at 23:58
  • Okay, I get it. However, if one considers the following splitting of the $ gl(5,\mathbb{R}) $-algebra : $ gl(5,\mathbb{R}) \approx \lbrace t \oplus \mathbb{R}^{4} \oplus gl(4,\mathbb{R}) \oplus \mathbb{R}^{4}{\ast} \rbrace r^{+} $, with $t$, $\mathbb{R}^{4}$, $\mathbb{R}^{4}{\ast}$ and $r^{+}$ being the algebras of time reflection with determinant +1, 4D translation group, graded 4D translation group and Abelian Lie subgroup with positive determinant, respectively. Then, is it true If we consider the aforementioned splitting as vector spaces? – Oussama Abdelghafour Dec 09 '23 at 09:18
  • Your notation there is a little unclear to me. In what sense are you multiplying by $r^+$? – Callum Dec 09 '23 at 13:11
  • I think it'a about a direct product in the version of groups not algebras, i.e. you have to transform all the algebras in the expression into their corresponding groups. Apart from this, I have a question: how do the direct and semi-direct products between groups transform when dealing with their associated algebras? – Oussama Abdelghafour Dec 09 '23 at 17:08
  • I'm afraid that doesn't make a whole lot of sense to me. Ignoring the $r^+$ part, that splitting is not a Killing orthogonal one. – Callum Dec 10 '23 at 15:29

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