# What is meant by the special linear group and special orthogonal group preserving orientation?

The special linear group is the group of matrices with determinant 1:

$$SL(n,\mathbb{R}) := \{ \mathbf{A} \in \mathbb{R}^{n \times n} \, | \, \det [\mathbf{A}] = 1 \}$$

The special orthogonal group, SO(n), is a subgroup with orthogonal matrices with determinant 1. This is also equivalent to the intersection of the special linear group and the orthognal group:

$$SO(n) := \{ \mathbf{A} \in \mathbb{R}^{n \times n} \, | \mathbf{A}^T\mathbf{A} = \mathbf{A}\mathbf{A}^T = \mathbf{I}, \, \det [\mathbf{A}] = 1 \}$$

These group are both "volume" preserving and "orientation" preserving. I read this here, and this is also stated on the Wikipedia articles. The fact that the groups contain volume preserving transformations make sense to me because the determinant is 1.

However, why are the groups with determinant 1 (e.g., SO(n) and SL(n,$$\mathbb{R}$$)) orientation preserving?

For example, SO(n) is just the group of rotation matrices, but rotation matrices don't preserve orientation (because they rotate...). I suppose, I don't understand what is implied by "orientation preserving".

• Are you familiar with the idea of right-handed and left-handed coordinate systems? Sep 21, 2020 at 20:31
• @eyeballfrog yes. Is that what is meant? Never mind. I see the answers below. Sep 21, 2020 at 20:40

Orientation is a concept meant for ordered bases of a vector space. If we have the ordered basis $$(b_1,\dots, b_n)$$ we say that it has a certain orientation. And we say that if we exchange two of the basis vectors to get, for instance, $$(b_k,b_2,\dots,b_{k-1},b_1,b_{k+1},\dots,b_n)$$, then we change the orientation of the basis. If we switch two more basis elements, then we change back to the original orientation (though not necessarily to the original ordered base). This way, we can define two orientations for an ordering of a given base. One we call left-handed, the other right-handed.

Now the determinant has this concept of orientation baked into its definition. Part of its definition is that if we exchange two columns of a matrix, then we flip the sign of its determinant (more formally, it is antisymmetric). What's also baked into it is that if a linear map with matrix representation $$M$$ (with respect to the ordered base above) is injective, then its determinant is non-zero. So if we take any matrix with non-zero determinant, then it maps an ordered base to a different ordered base: $$(b_1,\dots,b_n)\mapsto (Mb_1,\dots,Mb_n)$$. Now we said earlier that, for instance, the ordered base $$(Mb_k,Mb_2,\dots,Mb_{k-1},Mb_1,Mb_{k+1},\dots,Mb_n)$$ should have the opposite orientation of $$(Mb_1,\dots,Mb_n)$$, since we exchanged two of the basis vectors. And it miraculously (or not so miraculously after all ;) ) happens that we get to this other ordered base by exchanging two columns of $$M$$ and applying the new matrix to $$(b_1,\dots, b_n)$$. But at the same time we defined that the determinant of the matrix flips signs if we do that! So the sign of the determinant is a good way to define an orientation for all ordered bases: Let $$R:=(b_1,\dots,b_n)$$ be an ordered base , our reference base, which we define to be right-handed. Then we say that any other ordered base $$B$$ is right-handed if the determinant of the matrix transforming $$R$$ to $$B$$ is positive, and left-handed if the determinant is negative.

So to summarize: The sign of a matrix is the definition of what orientation preserving means. Positive sign is orientation preserving, negative sign is orientation inverting.

An orientation on a finite-dimensional real vector space $$V$$ of dimension $$d$$ is a choice of one of the two connected components of the top exterior power minus zero $$\wedge^d(V) \setminus \{ 0 \} \cong \mathbb{R} \setminus \{ 0 \}$$. This choice tells you which bases $$v_1, \dots v_d$$ of $$V$$ are "positively" versus "negatively" oriented with respect to the orientation ("right-" or "left-handed," except I don't know which should be which): positive means $$v_1 \wedge \dots \wedge v_d$$ lies in the connected component you picked, and negative means it lies in the other one.

The subgroup of $$GL(V)$$ preserving orientation (either orientation) is the subgroup $$GL^{+}(V)$$ of matrices with positive determinant, which is precisely the connected component of the identity. In particular, all rotations preserve orientation.

These groups are path-connected. If $$\gamma\colon [0,1]\to G$$ is a path from $$I$$ to an arbitrary element $$A$$, then $$\gamma(t)$$ applied to the standard basis is always a basis and continuously changes from the standard basis to $$A$$ applied to it. You cannot jump from a positively oriented to a negatively oriented basis along this path.