When we study differential geometry, we are particularly interested in geometric properties that are independent of change of coordinates. The same is true for physics. When we develop a physical theory, we require certain physical properties to be invariant under change of coordinates. Mathematically, change of coordinates (change of basis of a vector space) is given by linear isomorphisms of a vector space (a physical space) over a field $F$. Physicists usually call such invariance *symmetry*. The set of all linear isomorphisms of a vector space $V$ of dimension $n$ with a positive definite inner product is denoted by $\mathrm{GL}(n,F)$. $\mathrm{GL}(n,F)$ is a Lie group with function composition $\circ$. The Lie group $\mathrm{GL}(n,F)$ is called the *general linear group* of $V$. The dimension of $\mathrm{GL}(n,F)$ is $n^2$. The set of orientation preserving isomorphisms of $V$ forms a Lie subgroup of $\mathrm{GL}(n,F)$ and is denoted by $\mathrm{SL}(n,F)$. $\mathrm{SL}(n,F)$ is called the *special linear group* of $V$. In fact, a linear isomorphism $A\in\mathrm{SL}(n,F)$ if and only if $\det(A)=1$. The set of isometries (distance or length preserving linear isomorphisms) of $V$ forms a Lie subgroup of $\mathrm{GL}(n,F)$ and is denoted by $\mathrm {O}(n,F)$. $\mathrm {O}(n,F)$ is called the *orthogonal group* and it coincides with the set of all $n\times n$ orthogonal matrices, i.e. $$\mathrm {O}(n,F)=\{A\in\mathrm{GL}(n,F): AA^t=I\}.$$ In particuluar, orthogonal matrices whose determinants are equal to 1 preserve the orientation, and they form a Lie subgroup of $\mathrm{O}(n,F)$ called the *special orthogonal group* and it is denoted by $\mathrm{SO}(n,F)$.

Symmetry plays an important role in quantum physics. For instance, we expect that the angular momentum is conserved in quantum mechanics. The angular momentum is conserved if the hamiltonian $H$ is invariant under rotations in 3-space $\mathbb R^3$. Here $H$ being rotationally invariant means that $[R,H]=0$ or $RH=HR$, where $R$ is a rotation. Suppose that a wave function $\psi$ satisfies the Schrodinger equation $H\psi=E\psi$. If $H$ is rotationally invariant, then a rotated wave function $R\psi$ also satisfies the Schrodinger equation. Rotations in $\mathbb R^3$ form the orthogonal group $\mathrm{SO}(3)$. Here we have a problem. Rotations in $\mathbb R^3$ are $3\times 3$ orthogonal matrices while $\psi$ is a complex-valued map. So mathematically a rotation $R\in\mathrm{SO}(3)$ would not act on $\psi$. (The Lie group action would take place by a matrix multiplication.) For example, let us consider a two-state quantum mechanical system such as spin-up and spin-down states of an electron. In that case, the hamiltonian would be a $2\times 2$ complex matrix and states (wave functions) are $\mathbb C^2$-valued maps. In order for a matrix $R$ to act on $\psi$, it needs to be a $2\times 2$ complex matrix! It turns out that there are $2\times 2$ complex matrices that may represent rotations. The rotation group $\mathrm{SO}(3)$ is topologically not simply connected. This means that not all the loops (closed curves) in $\mathrm{SO}(3)$ can be continuously shrink to a point. The universal cover of $\mathrm{SO}(3)$ is $\mathrm{SU}(2)$ and that the covering map $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a 2:1 map. This means that $\mathrm{SU}(2)$ is a simply connected space that consists of two pieces each of which is identified with $\mathrm{SO}(3)$. $\mathrm{SU}(2)$ is called the *special unitary group* and is defined by $$\mathrm{SU}(2)=\{U\in\mathrm{SL}(2,\mathbb C): UU^\ast=I\}.$$ The covering map $\rho$ being a 2:1 map plays an important role in Dirac's theory of electrons. That means rotating an electron by $360^\circ$ does not bring back the original state of the electron (spin-up or spin-down). One needs to rotate the electron by $720^\circ$ in order to bring back the original state of the electron. If $F:\mathbb R^3\longrightarrow\mathrm{SO}(3)$ is a frame in $\mathbb R^3$, then we can lift this frame to $\mathrm{SU}(2)$. The map $\tilde{F}:\mathbb R^3\longrightarrow\mathrm{SU}(2)$ is called a lifted frame of $F:\mathbb R^3\longrightarrow\mathrm{SO}(3)$ if $F=\rho\circ \tilde{F}$. This relatioship can be simply represented by the following diagram:

$$\begin{array}{ccc}

& & \mathrm{SU}(2)\\

& \nearrow &\downarrow\\

\mathbb R^3 & \longrightarrow & \mathrm {SO}(3)

\end{array}$$

If we assume that the hamiltonian $H$ is time-independent, the Schrodinger equation $H\psi=E\psi$ or $i\frac{\partial\psi}{\partial t}=H\psi$ has solution $$\psi(\mathbf r,t)=U(t)\psi(\mathbf r,0),$$ where $U(t)=\exp\left(-\frac{i}{\hbar}Ht\right)$. As mentioned earlier, we want $U(t)$ to be a special unitary matrix, i.e. $U(t)\in\mathrm{SU}(2)$. As well-known in differential geometry, in order for $U(t)$ to be a special unitary matrix, the matrix $-iH$ needs to be a trace-free skew-hermitian matrix i.e. $-iH$ satisfies $$-iH+(-iH)^\ast=0$$ or equivalently $H=H^\ast$ and that $\mathrm{tr}H=0$. That is, the hamiltonian $H$ is hermitian! The space of all $2\times 2$ skew hermitian matrices is denoted by $\mathfrak{su}(2)$. It is the Lie algebra of the Lie group $\mathrm{SU}(2)$ and the exponential map $$\exp: \mathfrak{su}(2)\longrightarrow\mathrm{SU}(2);\ A\longmapsto\exp(A)$$ relates a Lie algebra element to a Lie group element. More generally, if we assume that the hamiltonian $H$ is invariant under isometries of $\mathbb R^3$ i.e. $[H,R]=0$ for $R\in\mathrm{U}(2)$, then the trace-free condition can be omitted and $H$ is required to be just hermitian since the Lie algebra $\mathfrak{u}(2)$ is simply the set of all $2\times 2$ skew-symmetric matrices. We learn in quantum mechanics that we require the hamiltonian $H$ to be hermitian in order to ensure that its eigenvalues (energies) are real. I have given another reason why $H$ is required to be hermitian from a geometric point of view. In order for the angular momentum to be conserved in quantum mechanics, $H$ must be required to be hermitian. In general, $U(t)=\exp\left(-\frac{i}{\hbar}Ht\right)$ is a unitary transformation on the Hilbert space $\mathcal H$ of states with hermitian hamiltonian $H$. The matrix corresponding to the unitary transformation $U(t)$ is called S (Scattering) matrix in quantum mechanics.

Note that states $\psi$ and $e^{i\theta}\psi$ have the same probability density, so they are not physically distinguishable. This means that the possible states are points in the complex projective space $\mathbb CP^1\cong S^2$. This can be seen as follows. Since we do not distinguish $\psi$ and $e^{i\theta}\psi$, distinct states are represented by elements (orbits) in the Riemannian symmetric space $\mathrm{U}(2)/\mathrm{U}(1)=\mathrm{SU}(2)/\mathrm{U}(1)\cong S^2$. Knowing that $\mathrm{SU}(2)=S^3$ and $\mathrm{U}(1)=S^1$, we have an interesting *exponential identity* (relation) $S^3/S^1=S^2$. In general, the following relations hold for $\mathrm{U}(n)$ and $\mathrm{SU}(n)$: $$\mathrm{U}(n+1)/\mathrm{U}(n)=\mathrm{SU}(n+1)/\mathrm{SU}(n)=S^{2n-1}.$$