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Show that special orthogonal matrix from $SO(n, \mathbb{R})$ has $\frac{(n-1)n}{2}$ independent real parameters.

I assume that this will be related to Euler angles somehow or specifically to its generalization. But how do we find degrees of freedom of such matrices? How do they look like?

Any hints would be helpful.

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An $n\times n$ real matrix contains $n^2$ real parameters. The column matrices of a real orthogonal matrix are normal and orthogonal to each other. There exist $n$ real matrix constraints for the normalization and $n(n-1)/2$ real constraints for the orthogonality. Thus, the number of independent real parameters for characterizing the elements of the groups $SO(N)$ is equal to $$ n^2-(n+n(n-1)/2)) = n(n-1)/2. $$

For the dimension as a Lie group, we can also just determine the dimension of its Lie algebra $\mathfrak{so}(n)$, consisting of skew-symmetric matrices, i.e., $n(n-1)/2$.

Edit: As remarked in the comments, this is not meant to be a proof. A correct proof requires much more theory.

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  • $\begingroup$ Could you elaborate on what "n real matrix constraints for the normalization and $n(n-1)/2$ real constraints for the orthogonality" means? $\endgroup$ – janusz Mar 5 at 20:23
  • $\begingroup$ Dietrich; your first proof is not a proof (and you kow that!). You make the same reasoning as most students: "since $A^TA$ is symmetric, the equality $A^TA=I_n$ is equivalent to $n(n+1)/2$ relations, that is the degree of freedom of an orthogonal matrix is $n(n-1)/2$"; that does not prove the algebraic independence of these relations. Moreover,I think that the OP does not even know what are algebraically independent relations. For years, I have been against this type of reasoning (which is only intuitive). I am surprised that a "maestro" like you presents it as a proof in its own right. $\endgroup$ – loup blanc Mar 6 at 15:13
  • $\begingroup$ @loupblanc you are right, this is not what I consider a proof. In fact it comes from a book and I just thought it could be taken as a plausible argument. As you say, the OP would not like more than this, I suppose. I would prove it via the Lie algebra. But also this needs more details, see the other answer. $\endgroup$ – Dietrich Burde Mar 6 at 16:30
  • $\begingroup$ Yes Dietrich, I agree with you. $\endgroup$ – loup blanc Mar 6 at 17:04
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Let

$\gamma(t) \in SO(n) \tag 1$

be a smooth path through the identity element $I$ of $SO(n)$:

$\gamma: I \to SO(n), \; \gamma(0) = I, \tag 2$

where

$I = (-\epsilon, \epsilon) \subset \Bbb R, \; \Bbb R \ni \epsilon > 0; \tag 3$

then we have

$\forall t \in I, \; \gamma^T(t) \gamma(t) = I, \tag 3$

by virtue of (2); if we differentiate this equation we obtain

$\dot \gamma^T(t) \gamma(t) + \gamma^T(t) \dot \gamma(t) = 0, \; \forall t \in I; \tag 4$

now taking

$t = 0, \; \gamma(t) = \gamma(0) = I, \tag 5$

we see that (4) becomes

$\dot \gamma^T(0) + \dot \gamma(0) = \dot \gamma^T(0) I + I \dot\gamma(0) = \dot \gamma^T(0) \gamma(0) + \gamma^T(0) \dot \gamma(0) = 0, \tag 6$

which yields

$\dot \gamma^T(0) = -\dot \gamma(0); \tag 7$

it follows then that the tangent space $T_ISO(n)$ at $I$ consists of skew-symmetric $n \times n$ real matrices; but it is well-known, and easy to see, that the dimension of this set of matrices over $\Bbb R$ is $n(n - 1)/2$. Therefore if we can show that every $n \times n$ matrix $K$ with

$K^T = -K \tag 8$

is in $T_ISO(n)$, we may conclude that in fact

$\dim SO(n) = \dfrac{n(n - 1)}{2}; \tag 9$

to this end, we define the path

$\gamma(t) = e^{Kt}; \tag{10}$

then

$\dot \gamma(0) = Ke^{K(0)} = KI = K; \tag{11}$

so every skew-symmetric $K$ is tangent to some curve (1) at $I$; thus we have finalized our demonstration that (9) binds.

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