I was just wondering, probably totally trivial question, but say I have a $2d$ vector in $\mathbb{R}^2$ or $\mathbb{C}^2$ and the scalar product: Which are the possible unitary groups leaving the scalar product invariant? I suppose one has $\mathrm{U}(1)$, $\mathrm{(S)U}(2)$ and $\mathrm{(S)U}(1,1)$ but it would be nice to know if that is all.
Also, sketchwise, how to proceed to higher dimensions?
Cheers!
Any (nondegenerate) bilinear form is the sum of a symmetric and skew-symmetric one.
The real symmetric bilinear forms are classified by signature $(p,q)$. The standard form on $\mathbb{R}^{p,q}$ is
$$ x\cdot y=(x_1y_1+\cdots+x_py_p)-(x_{p+1}y_{p+1}+\cdots+ x_{p+q}y_{p+q}). $$
Every such form looks like the above in some basis. The corresponding "unitary groups" are called orthogonal groups, $O(p,q)$, with corresponding special orthogonal groups $SO(p,q)$ too.
Over $\mathbb{C}$, we have $-x_jy_j=(ix_j)(iy_j)$, so with a change of basis, every symmetric bilinear form is just the usual dot product $x_1y_1+\cdots+x_ny_n$.
There is a unique skew-symmetric form (up to change of basis) in every even dimension (and none in odd dimension), with corresponding groups $Sp(2n,\mathbb{R})$ and $Sp(2n,\mathbb{C})$ (many, if not a majority, of sources would use $n$ instead of $2n$ in these notations).
Over complex vector spaces (or even quaternionic vector spaces), there are also sesquilinear forms, which are linear in one argument and conjugate-linear in the other. (In physics, conjugate-linearity is in the first argument, but often in math it will be in the second argument.) Every sesquilinear form is the sum of a Hermitian and a skew-Hermitian one.
The Hermitian sesquilinear forms are again classified by signature, the standard form being
$$ \langle x,y\rangle = (\overline{x_1}y_1+\cdots+\overline{x_p}y_p)-(\overline{x_{p+1}}y_{p+1}+\cdots+\overline{x_{p+q}}y_{p+q}) $$
on $\mathbb{C}^{p,q}$ or $\mathbb{H}^{p,q}$. The corresponding unitary groups are called $U(p,q)$ and $Sp(p,q)$. The former have special unitary versions $SU(p,q)$, but the latter don't (there isn't an obvious determinant of a quaternionic matrix, and the non-obvious determinant already only takes values $1$).
You can take a look at Wikipedia's article on Classical groups, or find other resources by googling "classical groups," but be warned there are many complicated and conflicting conventions and definitions.
All of these classifications of forms are essentially proved by generalizing the Gram-Schmidt process to show that "standard bases" for the forms exist.