The compact symplectic group $Sp(n)$ is the group of length-preserving linear transformations of $n$-dimensional quaternionic space. In one line,

$$Sp(n)=\{A\in M_n(\mathbb H):A^*A=I\}.$$

This looks like the familiar definition $O(n)=\{A\in M_n(\mathbb R):A^TA=I\}$, except that the entries are quaternions and $A^*$ means transpose and quaternionic conjugate. Nothing in the definition is nonlinear: after replacing each quaternion by four real numbers, an element of $Sp(n)$ is an ordinary $4n\times4n$ real matrix satisfying ordinary polynomial equations.

The notation has one trap. This article is about the compact group $Sp(n)$, sometimes written $USp(2n)$. A different group, the real symplectic group $Sp(2n,\mathbb R)$, preserves a skew-symmetric bilinear form and is noncompact. The two are related, but they are not the same group.

A quaternion is

$$q=a+bi+cj+dk,\qquad a,b,c,d\in\mathbb R,$$

with multiplication determined by

$$i^2=j^2=k^2=ijk=-1.$$

Thus $ij=k$ but $ji=-k$: multiplication is associative but not commutative. Conjugation and length are

$$\bar q=a-bi-cj-dk,\qquad |q|^2=q\bar q=a^2+b^2+c^2+d^2.$$

So $mathbb H$ is just $mathbb R^4$ as a real vector space, equipped with extra multiplication. The identity $|pq|=|p||q|$ is why quaternionic matrices can have a useful notion of orthogonality.

For a column $v=(q_1,\ldots,q_n)^T\in\mathbb H^n$, define

$$\langle v,w\rangle_{\mathbb H}=v^*w=\sum_{r=1}^n\bar q_r w_r, \qquad \|v\|^2=\langle v,v\rangle_{\mathbb H}=\sum_r|q_r|^2.$$

The inner product is quaternion-valued. Its real part is the ordinary dot product on $mathbb R^{4n}$. Consequently $A^*A=I$ says exactly what it says for real orthogonal matrices: the columns of $A$ form an orthonormal basis, now over $mathbb H$.

Left multiplication by $q=a+bi+cj+dk$ is a real-linear map $mathbb H\to\mathbb H$. In the basis $(1,i,j,k)$ its matrix is

$$L_q= \begin{pmatrix} a&-b&-c&-d\\ b&a&-d&c\\ c&d&a&-b\\ d&-c&b&a \end{pmatrix}.$$

Direct multiplication gives $L_q^TL_q=|q|^2I_4$. Therefore every unit quaternion gives a $4\times4$ orthogonal matrix. This is already the whole group $Sp(1)$:

$$Sp(1)=\{q\in\mathbb H:|q|=1\}=S^3\cong SU(2).$$

There is a useful way to draw all four coordinates without pretending that $S^3$ fits in ordinary space. Write a unit quaternion as

$$q=\cos\theta+\mathbf u\sin\theta,$$

where $\mathbf u=u_1i+u_2j+u_3k$ is a unit imaginary quaternion. The sphere below contains the imaginary vector $(b,c,d)=\mathbf u\sin\theta$; the circular gauge on top carries the missing angle $\theta$ and hence the real coordinate $a=\cos\theta$. Together they specify the full quaternion. The faint back half of the sphere and the dashed part of the vector indicate depth. Drag horizontally to choose $\theta$ and vertically to turn the imaginary direction.

The sketch below shows the four real coordinate axes $1,i,j,k$ and their images under left multiplication by $q=\cos t+i\sin t$. What looks like one complex rotation is actually the same-angle rotation in two perpendicular planes: the $(1,i)$ plane and the $(j,k)$ plane. It runs automatically; drag horizontally to choose $t$ yourself.

Replace each entry $a_{rs}$ of a quaternionic matrix by its $4\times4$ block $L_{a_{rs}}$. This turns $A\in M_n(\mathbb H)$ into an ordinary real $4n\times4n$ matrix. Under this replacement,

$$A^*A=I \quad\Longrightarrow\quad A_{\mathbb R}^TA_{\mathbb R}=I_{4n}.$$

Hence $Sp(n)$ is a subgroup of $O(4n)$. In fact its determinant is always $+1$, so

$$Sp(n)\subset SO(4n).$$

But most rotations of $mathbb R^{4n}$ are not in $Sp(n)$. A quaternionic-linear map must also respect right multiplication by $i,j,k$. Regard these as three fixed real-linear maps $I,J,K$ on $mathbb R^{4n}$. They satisfy

$$I^2=J^2=K^2=-I_{4n},\qquad IJ=K=-JI.$$

Then there is an entirely real definition:

$$Sp(n)=\{A\in SO(4n):AI=IA,\ AJ=JA,\ AK=KA\}.$$

Because $K=IJ$, commuting with $I$ and $J$ already forces commuting with $K$. This characterization says: $Sp(n)$ consists of rotations that do not disturb the three linked quarter-turn structures carried by quaternionic space.

In the next sketch, a point represents a vector in one quaternionic coordinate. The colored arrows represent $v$, $Iv$, $Jv$, and $Kv$ after a two-dimensional projection. Move the pointer to choose $v$. The visible lengths can change under projection, but the algebraic rules tying the four vectors together do not.

The sketch is a projection, not a proof. The proof is matrix multiplication: $I,J,K$ are orthogonal, square to $-I$, anticommute, and send every nonzero $v$ to four mutually orthogonal real vectors $v,Iv,Jv,Kv$. Repeating this in $n$ quaternionic coordinate directions organizes $mathbb R^{4n}$ into $n$ such frames.

Every quaternion can be written uniquely as $q=z+w j$ with $z,w\in\mathbb C$. The rule

$$z+w j\longmapsto \begin{pmatrix}z&w\\-\bar w&\bar z\end{pmatrix}$$

turns quaternion multiplication into ordinary complex matrix multiplication. Applying it entry by entry embeds an $n\times n$ quaternionic matrix into a $2n\times2n$ complex matrix.

Let

$$\Omega=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix}.$$

Then the compact symplectic group can also be written

$$Sp(n)=\{A\in U(2n):A^T\Omega A=\Omega\}=U(2n)\cap Sp(2n,\mathbb C).$$

This explains both words in “compact symplectic.” The equation $A^*A=I$ is unitary and makes the group compact; the equation $A^T\Omega A=\Omega$ preserves a complex skew form and makes it symplectic. It also explains the alternative notation $USp(2n)$.

Be careful about the two transposes: $A^*$ includes complex conjugation and controls length, while $A^T$ does not and controls the skew form.

A Lie group is simultaneously a group and a smooth space, with smooth multiplication and inversion. Here the entries of $A$ provide coordinates in a real Euclidean space, and $A^*A=I$ is a finite list of polynomial equations. The solution set is closed and bounded, hence compact. It is also a smooth manifold; its tangent space reveals the local shape.

Take a smooth path $A(t)$ in $Sp(n)$ with $A(0)=I$ and velocity $X=A'(0)$. Differentiate $A(t)^*A(t)=I$ at $t=0$:

$$X^*+X=0.$$

Thus the Lie algebra is

$$\mathfrak{sp}(n)=\{X\in M_n(\mathbb H):X^*=-X\}.$$

These are the quaternionic skew-Hermitian matrices. Count real parameters:

• each diagonal entry must be purely imaginary, giving $3n$ parameters;
• each pair above/below the diagonal is determined by one arbitrary quaternion, giving $4\binom n2$ parameters.

Therefore

$$\dim Sp(n)=3n+4\binom n2=n(2n+1).$$

For comparison, $dim O(n)=n(n-1)/2$ and $dim U(n)=n^2$. The dimension is the number of independent infinitesimal motions near the identity—not the size of the matrices representing them.

Exponentiating a tangent matrix produces a group element:

$$X\in\mathfrak{sp}(n)\quad\Longrightarrow\quad e^X\in Sp(n).$$

Indeed $(e^X)^*=e^{X^*}=e^{-X}=(e^X)^{-1}$. This is the same matrix exponential familiar from linear differential equations.

$Sp(n)$ acts on the unit sphere $S^{4n-1}\subset\mathbb H^n$. It acts transitively: given unit vectors $v$ and $w$, quaternionic Gram–Schmidt extends each to an orthonormal basis, and the matrix sending one basis to the other lies in $Sp(n)$.

Once the first column is chosen, the remaining columns form an element of $Sp(n-1)$ on its orthogonal complement. This gives a bundle

$$Sp(n-1)\longrightarrow Sp(n)\longrightarrow S^{4n-1}$$

and the homogeneous-space identity

$$S^{4n-1}\cong Sp(n)/Sp(n-1).$$

The dimension check is reassuring:

$$\dim Sp(n)-\dim Sp(n-1)=4n-1=\dim S^{4n-1}.$$

The sketch visualizes the real analogue of this column-by-column construction. Drag the yellow first column; the blue column is forced to stay perpendicular. Quaternionic Gram–Schmidt uses the same projection-and-subtraction recipe, but one chosen quaternionic column accounts for four linked real directions.

Iterating the bundle suggests—but does not by itself prove—that $Sp(n)$ is connected and simply connected. Both statements are true. Its Lie algebra belongs to the $C_n$ family, one of the four infinite classical families $A_n,B_n,C_n,D_n$.

As already seen,

$$Sp(1)=S^3=\{a+bi+cj+dk:a^2+b^2+c^2+d^2=1\}\cong SU(2).$$

It double-covers $SO(3)$. A unit quaternion $q$ rotates an imaginary quaternion $x=xi+yj+zk$ by

$$x\longmapsto qxq^{-1}.$$

The quaternions $q$ and $-q$ give the same rotation, which accounts for the two-to-one map.

$Sp(2)$ has dimension $2(5)=10$ and acts on $mathbb H^2\cong\mathbb R^8$. There is an exceptional isomorphism

$$Sp(2)/\{\pm I\}\cong SO(5),$$

or, at the simply connected level, $Sp(2)\cong Spin(5)$. This is a low-dimensional coincidence, not a pattern continuing for all $n$.

Real projective space identifies nonzero real vectors that differ by a real scale. Complex projective space does the same with complex scales. Quaternionic projective space is

$$\mathbb HP^{n-1}=(\mathbb H^n\setminus\{0\})/\mathbb H^\times.$$

After normalizing length, only unit-quaternion scalars remain, so there is a quaternionic Hopf fibration

$$Sp(1)=S^3\longrightarrow S^{4n-1}\longrightarrow\mathbb HP^{n-1}.$$

The group $Sp(n)$ moves quaternionic lines transitively. The line through the first basis vector is fixed by $Sp(1)\times Sp(n-1)$, giving

$$\mathbb HP^{n-1}\cong \frac{Sp(n)}{Sp(1)\times Sp(n-1)}.$$

For $n=2$, this says $\mathbb HP^1\cong S^4$. It is the quaternionic sibling of $\mathbb CP^1\cong S^2$ and connects $Sp(n)$ directly to the Hopf fibration.

Commutativity returns inside a useful subgroup. Choose the same complex plane $\{a+bi\}\subset\mathbb H$ in every diagonal entry. Then

$$T=\left\{\operatorname{diag}(e^{i\theta_1},\ldots,e^{i\theta_n})\right\}\cong(S^1)^n$$

is a maximal torus, so $Sp(n)$ has rank $n$. Every element of $Sp(n)$ is conjugate to an element of this torus: after a suitable quaternionic change of orthonormal basis, its essential motion is described by $n$ angles.

Relative to angle coordinates $e_1,\ldots,e_n$, the roots are

$$\pm e_r\pm e_s\quad(r\ne s),\qquad \pm2e_r.$$

This is the root system $C_n$, which is why the Lie algebra is written $\mathfrak{sp}(n)$ and classified as type $C_n$. The short roots mix two quaternionic coordinate directions; the long roots act within one. Counting roots gives $2n^2$, and adding the $n$-dimensional torus recovers

$$2n^2+n=n(2n+1)=\dim Sp(n).$$

The definitions can now be read as different views of one object:

The quaternionic row is the shortest definition. The complex row explains the word symplectic. The real row makes clear that the group is accessible using only real vectors and matrices: it is a specially constrained family of rotations in $4n$ dimensions.

Given a candidate transformation, any one of these routes is enough:

• Quaternionic form: verify $A^*A=I$.
• Complex form: verify both $A^*A=I$ and $A^T\Omega A=\Omega$.
• Real form: verify $A^TA=I$ and that $A$ commutes with the fixed matrices $I$ and $J$.
• Infinitesimal form: for a generator $X$, verify $X^*=-X$; then $e^{tX}\in Sp(n)$ for every real $t$.

The mental model to keep is simple: $O(n)$ moves real orthonormal frames, $U(n)$ moves complex orthonormal frames, and $Sp(n)$ moves quaternionic orthonormal frames. Each extra number system adds structure that a transformation must respect—and creates a richer group of coordinated rotations when viewed over the real numbers.