Basic question about dimensionality of Euclidean group I have a basic question about the dimensionality of the Euclidean group.
Why are degrees of freedom greater than the dimension? I thought that a degree of freedom is the same as a dimension, as in, $x\;\text{-}\;y$ plane is of dimension $2$ and therefore has $2$ degrees of freedom. Right?
So I don't understand how the degrees of freedom of the Euclidean group $E(n)$, defined by:
$$\dfrac{n(n+1)}{2}$$
turn out to be greater than the dimension $n$.  
For example, for dimension $n=2,$ the degrees of freedom are $3$, and for dimension $n=6$, the degrees of freedom are $n=3$. 
I don't understand this. Please explain. 
Thanks. 

First of all, each Euclidean transformation (an element of $E(n)$) has the form $Ax+b$ where $A\in O(n)$ (an orthogonal matrix) and $b\in R^n$. Clearly, $b$ depends on $n$ parameteris and we just have to compute the dimension of $O(n)$. Orthogonality of a matrix means that each column is a unit vector and any two distinct columns are orthogonal. Thus, you have $n^2$ variables (entries of the matrix) and $\frac{n(n+1)}{2}$ equations. By subtracting, this leaves you with  $\frac{n(n-1)}{2}$ parameters for the orthogonal group $O(n)$. (This argument sounds like cheating but there is a way to justify it using implicit function theorem.) Now, you get
$$
n+ \frac{n(n-1)}{2}= \frac{n(n+1)}{2},
$$ 
for the dimension of $E(n)$, which is the number you found somewhere. 
