What is the Dimension of SO(3) with Constraint det(O) = 1?

[shinobi20]

The group $\rm{O(3)}$ is the group of orthogonal $3 \times 3$ matrices with nine elements and dimension three which is constrained by the condition,

$$a_{ik}a_{kj} = \delta_{ij}$$

where $a_{ik}$ are elements of the matrix $\rm{A} \in O(3)$. This condition gives six constraints (can be worked out by brute force matrix multiplication to get the six equations) so it should have three degrees of freedom (d.o.f.) and therefore of dimension three, i.e. $\rm{d.o.f.} = 3 = 9-6$.

The group $\rm{SO(3)}$ is the same group $\rm{O(3)}$ but with additional constraint $\rm{det(O)} = 1$. Knowing that there is this additional constraint, shouldn't $\rm{SO(3)}$ have two d.o.f.? I know this is wrong but can anyone clarify this?

 

[PeroK]

The group $\rm{O(3)}$ is the group of orthogonal $3 \times 3$ matrices with nine elements and dimension three which is constrained by the condition,

$$a_{ik}a_{kj} = \delta_{ij}$$

where $a_{ik}$ are elements of the matrix $\rm{A} \in O(3)$. This condition gives six constraints (can be worked out by brute force matrix multiplication to get the six equations) so it should have three degrees of freedom (d.o.f.) and therefore of dimension three, i.e. $\rm{d.o.f.} = 3 = 9-6$.

The group $\rm{SO(3)}$ is the same group $\rm{O(3)}$ but with additional constraint $\rm{det(O)} = 1$. Knowing that there is this additional constraint, shouldn't $\rm{SO(3)}$ have two d.o.f.? I know this is wrong but can anyone clarify this?

Whether a constraint reduces the dimension depends on the constraint. $O(3)$ has two disconnected subsets, both of three dimensions (defined by determinant $\pm 1$). Constraining the determinant to $+1$ does not reduce the dimensionality.

A trivial example to illustrate this might be the real numbers. If we constrain $x > 0$, we still have a one dimensional line.

 

[shinobi20]

According to the discussion in this article (arxiv.org/abs/1312.3824), in section II third paragraph, $SU(2)$ has four parameters but the condition $\rm{det(U) = 1}$ gives one constraint. So I'm thinking the same for $SO(3)$.

 

[Infrared]

According to the discussion in this article (arxiv.org/abs/1312.3824), in section II third paragraph, $SU(2)$ has four parameters but the condition $\rm{det(U) = 1}$ gives one constraint. So I'm thinking the same for $SO(3)$.

The determinant of a unitary matrix can be any complex number of unit length, so specifying that it has unit determinant cuts down on one degree of freedom. But as @PeroK notes, the determinant of an orthogonal matrix can only be $\pm 1$, so specifying that an orthogonal matrix has determinant $1$ does not reduce the number of degrees of freedom.

 

[shinobi20]

The determinant of a unitary matrix can be any complex number of unit length, so specifying that it has unit determinant cuts down on one degree of freedom. But as @PeroK notes, the determinant of an orthogonal matrix can only be $\pm 1$, so specifying that an orthogonal matrix has determinant $1$ does not reduce the number of degrees of freedom.

Oh, so in the case of $\rm{O(3)}$, it is a fact that $det(\rm{O}) = \pm 1$, but for $\rm{U(2)}$ it is just a unitary group of dimension two, we need to specify the condition $det(\rm{U}) = 1$ which gives $\rm{SU(2)}$. So the keyword here is to specify (in the case of $\rm{U(2)}$)?

 

[PeroK]

Oh, so in the case of $\rm{O(3)}$, it is a fact that $det(\rm{O}) = \pm 1$, but for $\rm{U(2)}$ it is just a unitary group of dimension two, we need to specify the condition $det(\rm{U}) = 1$ which gives $\rm{SU(2)}$. So the keyword here is to specify (in the case of $\rm{U(2)}$)?

It's simpler than that. Here's an example:

First, take $z \in \mathbb C: \ |z| = 1$. There is one degree of freedom here: $z = e^{i\theta}$. If we further restrict to $z = 1$, then we lose the degree of freedom.

Second, take $x \in \mathbb R: \ |x| = 1$. There is no (continuous) degree of freedom here. If we further restrict to $x = 1$, then there is no continuous degree of freedom to lose.

 

[mathwonk]

in general, when you have a nice surjective mapping from X-->Y, then the dimension of X tends to equal the sum of the dimension of Y and the dimension of a general fiber of the map, i.e. of the inverse image of a general point of Y. in the case at hand the map is the determinant, the space X is a space of matrices, and the fiber is a subspace of those matrices with determinant one. In one case, X = U(2), the determinant takes as values all points of the circle, and the circle Y has dimension one. Hence the dimension of a general fiber, i.e. of the space SU(2) of those matrices in U(2) with determinant of value 1, has dimension one less than the dimension of U(2).

In the case of X = O(3), the determinant only takes as values the 2 possibilities {1,-1}, so this 2 point space Y has dimension zero. Hence the dimension of X = O(3), equals the dimension of the general fiber SO(3).

In the same way, if we started from a space of matrices whose determinant can be any non zero complex number, then restricting attention to those of determinant one would cut down the (real) dimension by 2.to state it in words, one has a family of objects, such as matrices, and one considers some quantity associated to those objects, such as the determinant. To compute the drop in dimension caused by restricting the value of that quantity to a particular value, such as seting the determinant equal to 1, ask yourself what is the possible range of values of that quantity? the drop in dimension equals exactly the dimension of the possible range in values. I.e. if the determinant can take a p dimensional range of values on all your original matrices, then setting it equal to 1 reduces the dimension of the space of matrices by p.

Thus since the determinant can take a one dimensional range of values on U(2), i,e, any complex number on the unit circle, then setting it equal to 1 reduces the dimension by one. Since on O(3) the determinant takes on a zero dimensional range of values, i.e. either 1 or -1, then setting it equal to 1 reduces the dimension by zero. In case of 2x2 invertible complex matrices, i.e. GL(2,C), since the determinant can take any non zero complex value, a range of values of (real) dimension two, the drop in dimension from setting the determinant equal to 1, i.e. from GL(2,C) to SL(2,C), is a drop of 2. I.e. since (real) dimension of GL(2,C) = 8, then dim.SL(2,C) = 6.