# Understanding some properties $U(2)$

I have read that the unitary group $U(2)$ can be naturally embedded as a subgroup of $SO(4)$. What exactly does this map look like? In particular, can we identify $U(2)$ as the set of matrices of $SO(4)$ that are $\textit{complex}$-linear?

Moreover, it seems that we still identify $\mathbb C^2$ with $\mathbb R^4$ here (correct?). Thus, if $U(2)$ is a subgroup of $SO(4)$, then wouldn't every unitary matrix be an orientation-preserving isometry of $\mathbb R^4$? But the problem with this is that this matrix could have determinant $-1$. How do we resolve this?

Finally (and a bit vaguely), if every unitary matrix is an isometry of $\mathbb R^4$, then how can we interpret geometrically the fact that this matrix is complex-linear? That is, can we view a unitary matrix as an isometry of $\mathbb C^2$ as a two-dimensional vector space over $\mathbb C$? (However, we would need a different norm in this case.)

• For your second question, can you think of a complex linear transformation with determinant $-1$...? ;) Generally (thinking of your third question), a complex linear transformation maps complex lines to complex lines, and (in an emphatic sense) most real $2$-planes are not complex lines. Commented Aug 9, 2017 at 11:46
• @AndrewD.Hwang I see, but are we still regarding the unitary map to be an isometry of $\mathbb R^4$ rather than of the $2$-dim. space $\mathbb C^2$? And if so, is there any sense in which we can talk about the map being an isometry of $\mathbb C^2$? Commented Aug 9, 2017 at 15:44
• In a word, yes: As a real vector space, $\mathbf{C}^{2}$ equipped with the standard Hermitian inner product is isometric to Euclidean $\mathbf{R}^{4}$. Commented Aug 9, 2017 at 15:54

To the first question: in general, we may embed $U(n)$ into $SO(2n)$ in several ways. One way would be $$A+iB\mapsto \begin{pmatrix} A & -B \cr B & A \end{pmatrix}.$$ Here the quotient $SO(2n)/U(n)$ is an irreducible symmetric space.

• How do we know that the determinant of this matrix is $1$? Commented Aug 10, 2017 at 3:47
• reduce to $\frac{1}{A^2+B^2}\left(\begin{array}{cc}A&-B\\B&A\end{array}\right)$ perhaps Commented Feb 6, 2023 at 1:22

Let $V$ be a complex vector space with inner product $\langle-,-\rangle_{\mathbb{C}}$. Since $\mathbb{R}\hookrightarrow\mathbb{C}$ we may also interpret $V$ as a real vector space, but with what real inner product? We could take an orthonormal basis $\mathcal{B}$ for $V$ as a complex vector space, and then use $\mathcal{B}\sqcup i\mathcal{B}$ as a basis for $V$ as a real vector space to give it a real inner product. The result in fact does not depend on which basis $\mathcal{B}$ we use, since the new real inner product will in fact be $\langle u,v\rangle_{\mathbb{R}}=\mathrm{Re}\langle u,v\rangle_{\mathbb{C}}$. Note that taking the real part does not actually lose information - we may recover $\langle -,-\rangle_{\mathbb{C}}$ from $\langle -,-\rangle_{\mathbb{R}}$ by writing $\langle u,v\rangle_{\mathbb{C}}=\alpha+\beta i$, noticing $\langle iu,v\rangle_{\mathbb{C}}=-\beta+i\alpha$ and concluding $\langle u,v\rangle_{\mathbb{C}}=\langle u,v\rangle_{\mathbb{R}}-i\langle iu,v\rangle_{\mathbb{R}}$.

Notice also that multiplication-by-$i$ on $V$ will be skew-Hermitian with respect to the real inner product structure, i.e. $\langle iu,v\rangle_{\mathbb{R}}=-\langle u,iv\rangle_{\mathbb{R}}$. Geometrically this should come as no surprise - any complex basis $\mathcal{B}$ allows us to decompose $V$ into a bunch of complex one-dimensional subspaces, which in turn means we may decompose $V$ into a bunch of real two-dimensional subspaces, and clearly multiplication-by-$i$ acts as a right-angle rotation in each such plane.

Conversely, given any real inner product space $V$ with a skew-symmetric isometry $X$ on it, we may turn $V$ into a complex inner product space by treating $X$ as multiplication-by-$i$. So, explicitly, that means we define $(a+bi)v:=av+bX(v)$. The group of $\mathbb{C}$-linear transformations $\mathrm{GL}_{\mathbb{C}}(V)$ will be a subgroup of the $\mathbb{R}$-linear ones, $\mathrm{GL}_{\mathbb{R}}(V)$. In fact, $\mathrm{GL}_{\mathbb{C}}(V)$ will almost by definition be the centralizer of $X$ within $\mathrm{GL}_{\mathbb{R}}(V)$, since being $\mathbb{R}$-linear and commuting with multiplication-by-$i$ (i.e. $X$) is equivalent to being $\mathbb{C}$-linear.

(1)

The group of $\mathbb{C}$-linear transformations which preserve the complex inner product, $\mathrm{U}(V)$, will be precisely the centralizer of $X$ within $\mathrm{O}(V)$ (the group of transformations which preserve the real inner product). To see why $\mathrm{U}(V)= C_{\mathrm{O}(V)}(X)$, prove that the following are equivalent:

• $g$ is $\mathbb{C}$-linear and preserves $\langle-,-\rangle_{\mathbb{C}}$
• $g$ is $\mathbb{R}$-linear, commutes with $X$, and preserves $\langle-,-\rangle_{\mathbb{R}}$

Concretely, with explicit matrices, if we have the coordinate vector space $\mathbb{R}^{2n}$ equipped with the dot product, and $X$ is the right-angle rotation of the subspace $\mathbb{R}^n\oplus0$ to become $0\oplus\mathbb{R}^n$ (given explicitly by $X(u,v)=(-v,u)$ under the identification $\mathbb{R}^{2n}=\mathbb{R}^n\oplus\mathbb{R}^n$), then we have a one-to-one homomorphism between matrix groups $\mathrm{U}(n)\to\mathrm{O}(2n)$ given by Dietrich in his answer.

(2)

It's true that an element of $\mathrm{U}(n)$ may have complex determinant $-1$ whereas all elements of $\mathrm{SO}(2n)$ have determinant $+1$, but the map $\mathrm{U}(n)\to\mathrm{SO}(2n)$ is not determinant preserving. Let's just look at the simplest case of $\mathbb{C} \,=" \mathbb{R}^2$: multiplication by $-1$ on $\mathbb{C}$ has complex determinant $-1$ (since $\det[z]=z$ for $1\times1$ matrices $[z]$), but on $\mathbb{R}^2$ it is a half-turn (i.e. $180^{\circ}$ rotation) which is represented by the matrix $-I_2$ and has determinant $(-1)(-1)=+1$.

In fact, the relationship between the two determinants is $\det_{\mathbb{R}}(g)=|\det_{\mathbb{C}}(g)|^2$, and since every $g\in\mathrm{U}(n)$ has complex determinant of modulus $1$, the corresponding elements of $\mathrm{O}(2n)$ have determinant $1$, so are in $\mathrm{SO}(2n)$. Alternatively, since $U(n)$ is connected and $U(n)\to\mathrm{O}(2n)$ is continuous, the image must be connected, which means it must lie in the connected component, which is $\mathrm{SO}(2n)$.

(3)

This is one I am not sure there is a satisfying answer for, short of being able to natively, directly visualize four dimensions with a better-than-human brain. However I can give geometric meaning to the transformations and explain an upgraded version.

Any rotation of $\mathbb{R}^m$ will be a bunch of 2D rotations in oriented, pairwise orthogonal planes (2D subspaces). I mentally picture it as a bunch of dials that can be turned independently.

The set of planes will in fact an invariant associated with the rotation ... except for a set of rotations of measure zero. Indeed, the set of planes is an invariant of the rotation if and only if all the angles of rotation are distinct up to orientation. At the opposite extreme lie isoclinic rotations, which is where all the angles of rotation are the same.

In the case of $\mathbb{R}^4$, isoclinic rotations are a pair of rotations in a pair of planes, which one may think of as a "twist" (like an Indian rug burn or purple nurple of your childhood). They come in two types, depending on whether the orientation of the two planes "add up" to the orientation of the whole 4D space or its opposite, called left and right isoclinic rotations respectively. Multiplication-by-$i$ is just one specific isoclinic rotation, and all of the other elements of $\mathrm{SU}(2)$ are all the isoclinic rotations of the other type.

Quaternions are useful for this. They are are a noncommutative 4D number system $\mathbb{H}$ whose elements may be regarded as pretend-sums of real scalars and 3D vectors in $\mathrm{span}\{i,j,k\}$. The coordinate vectors (in any ordered orthonormal basis with the correct orientation) are anticommuting (e.g. $ij=-ji$) and square to $-1$ (e.g. $i^2=-1$). They have a norm coming from using $1,i,j,k$ as a basis, and this norm is multiplicative. The unit quaternions form a $1\times1$ matrix group $\mathrm{Sp}(1)$.

Left and right multiplication by unit quaternions does not change the norm, so induces a map $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to\mathrm{SO}(4)$. This is in fact a double covering with kernel $(-1,-1)$. The left factor, corresponding to left multiplication by unit quaternions, corresponds to left isoclinic rotations, and the right factor to right isoclinic rotations. Since $\mathbb{H}$ is associative, left and right multiplications commute, so all left and right isoclinic rotations commute.

If we interpret $\mathbb{C}\subset\mathbb{H}$ and multiply $\mathbb{H}$ by complex scalars on the right (so, a right complex vector space), we may identify $\mathbb{H}\cong\mathbb{C}^2$, and in this case the $\mathbb{C}$-linear rotations are $\mathrm{Sp}(1)\times\mathrm{U}(1)\subset\mathrm{Sp}(1)\times\mathrm{Sp}(1)$. So, the fact that all left isoclinic rotations commute with all right isoclinic rotations is an upgraded version (since $\mathrm{U}(1)$ is a subset of $\mathrm{Sp}(1)$).

This is unique to 4D: all left (right) isoclinic rotations form a group under composition, and commute with all (right) left isoclinic rotations. It does not happen in higher dimensions. (There is no distinction between left/right in odd dimensions anyway.)