Do Carmo's Curves and Surfaces book states that "two ordered bases $e = \{e_i\}$ and $f = \{f_i\}$, $i = 1,\ldots,n$, of an $n$-dimensional vector space $V$ have the same orientation if the matrix of change of basis has positive determinant."

This same concept of "orientation" is the subject of another question on stack, but I am concerned with a purer intuitive understanding. What is the idea behind assigning this notion of same orientation to this particular circumstance with positive determinant? What properties of a matrix with positive determinant motivate this definition?

  • $\begingroup$ It is just because matrices respresenting orthonornal basis are elements of $O_n(\mathbb R)$ so the determinant of a matrix to change between two such basis is $+1$ or $-1$. $\endgroup$
    – Maxime
    Commented Jun 13 at 10:28
  • $\begingroup$ If you’ve done a bit of physics, you know that depending on your orientation, vector products might change signs. $\endgroup$
    – Maxime
    Commented Jun 13 at 10:30
  • $\begingroup$ Orthogonal matrices with determinant one are rotations. That group is denoted by $SO(n)\,.$ When the determinant is $-1$ they are reflections. Study a few low dimensional bases with that in mind. To consider non-orthonormal matrices with positive/negative determinant is the next minor generalization you should go through. $\endgroup$
    – Kurt G.
    Commented Jun 13 at 10:36
  • 1
    $\begingroup$ $GL_n(\mathbb R)$ has two path connected components, the subgroup with positive determinant and the 2nd coset with negative determinant $\endgroup$ Commented Jun 13 at 15:41
  • $\begingroup$ Possibly of interest: math.stackexchange.com/questions/1551768/… and/or math.stackexchange.com/questions/1884953/… $\endgroup$ Commented Jun 13 at 19:58

2 Answers 2


In general, the determinant of a matrix gives us information about the scale factor of 'volume' in our space. To make this more concrete, we consider the following image:

enter image description here

(I've called the angle between the $x$-axis and $a$ the angle $\alpha$ and similar for the angle for $b$.

Suppose we have $a,b \in \mathbb{R}^2$ spanning this paralellogram $P$ (dont pay attention to the numbers in the picture). If we want to find the area of this, we can find (by some trigonometry) that the area $A$ is given by: \begin{align} A = |\: \|a\| \|b\| \sin(\beta - \alpha) \:|. \end{align} If we want to write this as a function of the coordinates $(a_1,a_2) = a$ and $(b_1,b_2) = b$, then we may note that $$a_1 = \|a\| \cos(\alpha)$$ and $$a_2 = \|a\| \sin(\alpha).$$ Similarly $$b_1 = \|b\| \cos(\beta)$$ and $$b_2 = \|b\| \sin(\beta).$$

Then: \begin{align*} \pm A &= \|a \| \; \|b\| \sin(\beta-\alpha) \\ &=\|a\| \|b\| (\cos(\alpha) \sin(\beta) - \sin(\alpha) \cos(\beta)) \\ &= \|a \| \cos(\alpha) \|b\| \sin(\beta) - \|a\| \sin(\alpha) \|b\| \cos(\beta)\\ &= a_1 b_2 - a_2 b_1 \end{align*} Then $A = |a_1 b_2 - a_2 b_1|$. By taking the absolute value, we lose some information. Note that $a_1 b_2 - a_2 b_1$ is positive if $\sin(\beta-\alpha)>0$ and that it is negative when $\sin(\beta-\alpha)<0$. From this it follows that $\sin(\beta-\alpha)$ is positive if and only if $0 < \beta-\alpha<\pi$.

Geometrically speaking, that means that if we take a point $a$, and turn it over the shortest angle so it is parallel to $b$, then this would require us to turn anticlockwise. The sign of $a_1 b_2 - a_2 b_1$ thus gives us information about how the two vectors are oriented with respect to eachother. If we instead had to turn the other way, then $b$ would have to be "below" $a$ in the image. Now we can make the connection with the determinant:

Note that the determinant $\det \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} = a_1 b_2 - a_2 b_1$ which is exactly the 'signed area' we found before.

We call an ordered pair of vectors $a,b$ positively oriented if $\det(a,b) >0$ and negatively oriented if $\det(a,b)<0$. Note that is is of importance which of the two you pick as first one, and which is the second one.

For example, the standard basis $e_1,e_2$ is positively oriented, but the oriented pair $e_2,e_1$ is negatively oriented.

One can generalize this to higher dimensions, but the idea stays the same (in $3$ dimensions, you would not have a paralellogram, but a paralellopiped in the image instead).

Conclusion: If you have a basis, and you apply a change of bases which changes the orientation, the sign of the determinant will be negative. This means that the 'aboveness' and 'belowness' of the two vectors has been switched. If the determinant is positive, the orientation is preserved (altough volumes may be scaled).


Your right hand is a right hand no matter how you orient it in space. You can think of your thumb, index, and middle finger as an ordered basis for $\Bbb R^3$, or "frame" (assuming you don't contort your fingers to make them linearly dependent). The set of all frames may be topologized as a submanifold of $\Bbb R^3\times\Bbb R^3\times\Bbb R^3$. Indeed, viewing the vectors as columns of a matrix, frames correspond to invertible matrices, so we're talking about the space ${\rm GL}(3,\Bbb R)$ of invertible matrices. As you move and reorient your right hand in space, the corresponding matrix varies in ${\rm GL}(3,\Bbb R)$. The same goes for your left hand, but it lies in a different path-connected component. Thus, the orientations of a real vector space $V$ may be defined as the equivalence classes of ordered bases, or equivalently the connected components of ${\rm GL}(V)$.

In the 2D case, we can imagine circular arrows drawn on a surface differentiating clockwise vs. anticlockwise; these correspond to classes of ordered pairs of linearly independent vectors (whichever way the second vector is a convex angle from the first, the circular arrow is drawn in that direction).

The QR decomposition (via the Gram-Schmidt process, which allows us to orthonormalize any basis) tells us the multiplication map ${\rm O}(n)\times B\to{\rm GL}(n,\Bbb R)$ is a diffeomorphism, where $B$ is the subgroup of triangular matrices with positive diagonal entries. Note $B\simeq \Bbb R^{n(n+1)/2}$ is diffeomorphic to a Euclidean space, so ${\rm GL}(n,\Bbb R)$ deformation retracts onto ${\rm O}(n)$, and thus the components of ${\rm GL}(n,\Bbb R)$ correspond to those of ${\rm O}(n)$. What this means is that for real inner product spaces $V$ an equivalent definition of an orientation is an equivalence class of ordered orthonormal bases.

It is possible to show any matrix in ${\rm O}(n)$ is path-connected to one in ${\rm O}(n-1)$ for $n>1$, so by induction ${\rm O}(n)$ has at most as many components as ${\rm O}(1)=S^0\cong\Bbb Z_2$. (One can consider the long homotopy exact sequence for the fiber bundle ${\rm O}(n-1)\to{\rm O}(n)\to S^{n-1}$ for this conclusion. The elementary version of this is the path-lifting property for an orbit map ${\rm O}(n)\to S^{n-1}$. Equivalently, one may use an "arc" of a one-parameter subgroup $R$ of plane rotations to rotate the last column of a matrix $A$ to $e_n$, then apply $R(\theta)$ to $A$ pointwise.)

Consider the determinant $\det:{\rm O}(n)\to S^0=\{\pm1\}$. (The range can only contain $\pm1$, just apply $\det$ to $A^TA=I$ to see, and both values are realized, as $\det I=1$ and $\det F=-1$ for any reflection $F$.) Since $S^0$ is disconnected, its fibers must also be disconnected, so ${\rm O}(n)$ has exactly two components, corresponding to determinant $\pm1$. Moreover, elements of $B$ have positive determinant, so the components of ${\rm GL}(n,\Bbb R)$ correspond to matrices with positive/negative determinant.

For a manifold, an orientation is a choice of orientation of each tangent space which varies smoothly from point to point. For a Riemannian manifold, the tangent spaces are inner product spaces and the earlier point about those applies. A connected manifold either has zero or two orientations, in the latter case it is called orientable. For non-orientable manifolds, it is possible to pick an orientation for one tangent space, then move it around in space until it comes back to the opposite orientation at the original point (consider circular arrows drawn on a Mobius band).


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