1.2.7 Tangent Spaces

Lie matrix groups are smooth man­i­folds that tend to be curved ob­jects, but it turns out that the char­ac­ter­is­tic math­e­mat­ics of a matrix group can be en­cap­su­lat­ed in the flat space, that is, the tan­gent space $\mathfrak{T}_{\epsilon}(G)$ The tan­gent space of a group $G$ con­sists of smooth paths $A(t)$ moving through the iden­ti­ty $\epsilon$. Smooth de­fines that de­riv­a­tives of these paths exist $\left( A^{'}(t) \right)$ and if $A(0) = \epsilon$, then $A^{'}(0)$ is called the tan­gent vector of $A(t)$ at $\epsilon$.¹ De­ter­min­ing the matrix de­riv­a­tive of $A$ cen­tered at the iden­ti­ty $A^{'}(0)$ to eval­u­ate the smooth paths through is straight­for­ward for the gen­er­al and spe­cial linear groups $GL\left( n,\mathbb{C} \right)$ and $SL\left( n,\mathbb{C} \right).$The tan­gent vec­tors of these groups will take the form of a $n \times n$ skew-sym­met­ric matrix $X,$de­fined for $X\in \mathbb{C}$, as

For $X$, a square Lie al­ge­bra matrix, the ex­po­nen­tial map­ping $e^{X}$ yields a smooth path via dif­fer­en­tia­bil­i­ty, with a tan­gent vector at the iden­ti­ty. For the group $G,$we set a smooth path $A(t)$ to rep­re­sent

and by being smooth, there exists a de­riv­a­tive of $A(t)$ that has the fol­low­ing form

Eval­u­at­ing the paths through the iden­ti­ty means con­sid­er­ing $A^{'}(t)$ when $t = 0$ , which yields the tan­gent vec­tors that com­prise the Lie al­ge­bra $X$

For the sake of clar­i­ty, the op­er­a­tor no­ta­tion in the fol­low­ing the­o­rem acting on $X$, is the fol­low­ing: the trans­pose is de­not­ed as $X^{T}$, the con­ju­gate as $\overline{X}$, and the con­ju­gate trans­pose is $X^{\dagger}$ and is equiv­a­lent to ${\overline{X}}^{T}.$

The­o­rem (1.2.145) – The tan­gent vec­tors $X$ at $\epsilon$, for the $O(n),\ U(n),\ Sp(n)$ groups are ma­tri­ces of the fol­low­ing forms:

• $\mathfrak{T}_{\epsilon}\left( O(n) \right) = \{ X \in G_{n}\left( \mathbb{R} \right):X + X^{T} = 0\}$.

Proof – The ma­tri­ces $A \in O(n),$ by way of their or­thog­o­nal­i­ty, sat­is­fy $AA^{T} = \epsilon.$ For $A = A(t),$ rep­re­sent­ing a smooth path that orig­i­nates at the iden­ti­ty $\epsilon$, we eval­u­ate the de­riv­a­tive of

The prod­uct rule holds for this de­riv­a­tive ex­pres­sion, given the iden­ti­ty, $\epsilon$, is a con­stant, and the de­riv­a­tive of a con­stant is zero, $\frac{d}{dt}(\epsilon) = 0$. Solv­ing by the prod­uct rule yields

and as both $A(0),\ A(0)^{T} = \epsilon$, for $t = 0$ the above be­comes

There­fore, any tan­gent vector $X = A^{'}(0)$ will sat­is­fy the skew-sym­me­try con­di­tion $X + X^{T} = 0.$

The ma­tri­ces $A \in U(n)$, by way of their uni­tary con­di­tions, sat­is­fy $AA^{\dagger} = \epsilon.$ For $A = A(t)$, a smooth path that orig­i­nates at the iden­ti­ty $\epsilon$, we eval­u­ate the de­riv­a­tive of

The prod­uct rule holds for this de­riv­a­tive ex­pres­sion, given the iden­ti­ty, $\epsilon$, is a con­stant, and the de­riv­a­tive of a con­stant is zero, $\frac{d}{dt}(\epsilon) = 0$. Solv­ing by the prod­uct rule yields

and as both $A(0),A^{\dagger}(0) = \epsilon$, for $t = 0$, the above be­comes

There­fore, any tan­gent vector, $X = A^{'}(0)$, will sat­is­fy the skew-sym­me­try con­di­tion, $X + X^{\dagger} = 0.$ The ma­tri­ces $A \in Sp(n)$, by way of their quater­nion con­di­tions, sat­is­fy $AA^{\dagger} = \epsilon.$ For $A = A(t)$, a smooth path that orig­i­nates at the iden­ti­ty, $\epsilon$, we eval­u­ate the de­riv­a­tive of

The prod­uct rule holds for this de­riv­a­tive ex­pres­sion, given the iden­ti­ty, $\epsilon$, is a con­stant, and the de­riv­a­tive of a con­stant is zero, $\frac{d}{dt}(\epsilon) = 0$. Solv­ing by the prod­uct rule yields

and as both $A(0),A^{\dagger}(0) = \epsilon$, for $t = 0$, the above be­comes

There­fore, any tan­gent vector $X = A^{'}(0)$, will sat­is­fy the skew-sym­me­try con­di­tion $X + X^{\dagger} = 0.$

$\square$

For the skew-sym­me­try con­di­tion $X + X^{T} = 0$, the el­e­ments along the di­ag­o­nals of $X$ will be 0, and the trans­pose of each el­e­ment in $X$ sat­is­fies $x_{j,k} =$-$x_{j,k}$. For the $X + X^{\dagger} = 0$ cases of the $U(n),Sp(n)$ groups, the ma­tri­ces are called skew-Her­mit­ian, where the matrix en­tries in $X$ sat­is­fy $x_{j,k} =$-${\overline{x}}_{j,k}$, and the di­ag­o­nal en­tries of $X$ are purely imag­i­nary. It should be noted that some tan­gent spaces at the iden­ti­ty can be shared for spe­cif­ic groups and their sub­groups; a case in point is the $O(n)$ groups and their $SO(n)$ sub­groups.

1. Still­well, John, "Naive Lie Theory", Springer, 2008