# What about groups from non-orthogonal matrices?

I am trying to characterize all finite groups which can be made from two 4D-real-matrix-represented generators. Though matrix coefficients must be real (I am avoiding complex coefficients since they effectively double the dimension to 8D), matrix columns are generally not orthogonal (but, through a transformation, one matrix could be considered orthogonal).

Given that the two matrices form a finite group, their determinants must be +1 or -1.

If columns were orthogonal for both matrices, I believe I would always be looking at one of the 6 possible polytopes (ignoring reflection-multiplied groups from lower-dimensional polytope groups, e.g., 2D cyclic groups which can be of any length).

Or, will I see more groups if I include non-orthogonal matrices? Conversely, if I see one of the 6 possible polytopes, will the two matrices be simultaneously orthonormally-reducible?

• Why the downvote? Maybe I'm missing something obvious to you, but at least give me a hint... Jun 22, 2021 at 23:35
• I'm not a downvoter, but the question is a bit unclear. Jun 23, 2021 at 2:06
• @Randall The title is unclear, but it was meant to draw attention to mathematicians/physicists always talking about "Special Orthogonal" SO(n) groups as if no one cares about non-orthogonal matrices, or as if they are equivalent somehow...are they? Jun 23, 2021 at 2:11

There is a basic trick in the representation theory of finite (or more generally, compact) groups that allows you to assume your matrices are orthogonal. Namely, given any finite subgroup $$G\subseteq GL_n(\mathbb{R})$$, you can always find an inner product which is preserved by each element of $$G$$ (so if you change to an orthonormal basis with respect to that inner product, $$G$$ will become a subgroup of the orthogonal group). The trick is that any positive linear combination of inner products is an inner product, so let $$\langle\cdot,\cdot\rangle$$ be any inner product and define $$\langle x,y\rangle'=\sum_{g\in G}\langle gx,gy\rangle.$$ Then $$\langle\cdot,\cdot\rangle'$$ is an inner product, and is $$G$$-invariant because applying an element of $$G$$ will just permute the summands.