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A rigid body moving in $\mathbb{R^2}$ has 3 degrees of freedom and in $\mathbb{R^3}$ has 6 degrees of freedom. Could you please help me show that a rigid body moving in $\mathbb{R^n}$ has $\frac{n+n^2}{2}$ degrees of freedom? And how many are translational and how many are rotational?

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The idea here is the following: the configuration manifold of a rigid body in $\mathbb{R}^n$ is $Q = \mathbb{R}^n\times SO(n)$. The idea behind this is that we only need the three coordinates of the center of mass plus specifying a frame fixed at that point. The $\mathbb{R}^n$ part locates the center of mass and the $SO(n)$ part gives the frame.

Another way to look at it is that we are specifying a rigid motion. So $\mathbb{R}^n$ gives the translation and $SO(n)$ the rotation to go from a well-known initial configuration to the current one.

Now the number of degrees of freedom is just the dimension of $Q$. As it's known from differential geometry this will be the sum of the dimensions

$$\dim Q = \dim \mathbb{R}^n + \dim SO(n),$$

but $\dim \mathbb{R}^n=n$, while it's also known that $\dim SO(n) = \frac{n(n-1)}{2}$, thus

$$\dim Q = n + \dfrac{n(n-1)}{2}=\dfrac{2n + n^2 - n}{2} = \dfrac{n+n^2}{2}.$$

Also it's clear that the translational degrees of freedom are the ones associated with $\mathbb{R}^n$ and the rotational ones are those associated with $SO(n)$.

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