Tangent Spaces

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Tangent spaces

First definition using curves

Let $M$ be an $m$-dimensional manifold. For $p \in M$, the tangent space $T_p M$ is the set of all linear maps $v : C^\infty(M) \to \R$ of the form \(v(f) = \frac{d}{dt}\bigg\vert_{t=0} f(\gamma(t))\) where $\gamma \in C^\infty(J, M)$ is a smooth map with $\gamma(0) = p$ and $J \subseteq \R$ is an interval containing $0$. The elements of $T_p M$ are called tangent vectors to $M$ at $p$.

Charts and tangent spaces

Let $(U, \phi)$ be a coordinate chart around $p$. A linear map $v : C^\infty(M) \to \R$ belongs to $T_p M$ if and only if it has the form \(v(f) = \sum_{i=1}^m a^i \frac{\partial (f \circ \inv\phi)}{\partial u^i} \bigg\vert_{\u=\phi(p)} = \a \cdot \nabla (f \circ \inv \phi)\vert_{\phi(p)}\) for some $\a = (a^1, …, a^m) \in \R^m$.

As a consequence, $T_p M$ is an $m$-dimensional vector space.

Proof.

($\Rightarrow$) Suppose $v \in T_p M$, then for any $f \in \C^\infty(M)$,

\[v(f) = \frac{d}{dt} \bigg\vert_{t=0} f(\gamma(t))\]

for some $\gamma \in C^\infty(J, M)$ with $\gamma(0) = p$. Inserting $\phi$, we see that

\[v(f) = \frac{d}{dt} \bigg\vert_{t=0} f(\inv \phi \circ \phi \circ \gamma(t))\]

so if $\widetilde \gamma = \phi \circ \gamma$

\[v(f) = \frac{d}{dt} \bigg\vert_{t=0} (f \circ \inv \phi)(\widetilde \gamma(t))\]

Since $\widetilde \gamma : J \to \R^m$, we have a composition of functions between Euclidean spaces, so we can use the chain rule to see

\[\begin{align*} v(f) &= \sum_{i=1}^m \frac{\partial (f \circ \inv \phi)}{\partial u^i} \bigg\vert_{\u=\widetilde\gamma(0)} \frac{\partial \widetilde\gamma^i}{\partial t} \bigg\vert_{t=0} \\ &= \sum_{i=1}^m \frac{\partial (f \circ \inv \phi)}{\partial u^i} \bigg\vert_{\u=\phi(p)} a^i \\ &= \nabla (f \circ \inv \phi)\vert_{\phi(p)} \cdot \a \end{align*}\]

where $\ds a^i = \frac{d\widetilde\gamma^i}{dt}\bigg\vert_{t=0}$, so $\ds \a = \frac{\partial \widetilde \gamma}{\partial t} \bigg\vert_{t=0}$. So we have shown what we wanted.

($\Leftarrow$) Suppose we have a map

\[v(f) = v(f) = \sum_{i=1}^m a^i \frac{\partial (f \circ \inv\phi)}{\partial u^i} \bigg\vert_{\u=\phi(p)}\]

for some $\a = (a^1, …, a^m)$. Then we want to find a curve $\gamma$ so that

\[v(f) = \frac{d}{dt} \bigg\vert_{t=0} f(\gamma(t))\]

To do this, let $\gamma(t) = \inv\phi(\phi(p) + t\a)$. Then,

\[\begin{align*} \frac{d}{dt} \bigg\vert_{t=0} f(\gamma(t)) &= \frac{d}{dt} \bigg\vert_{t=0} f(\inv\phi(\phi(p) + t\a)) \\ &= \frac{d}{dt} \bigg\vert_{t=0} (f \circ \inv\phi)(\phi(p) + t\a) \\ &= \sum_{i=1}^m a^i \frac{\partial (f \circ \inv\phi)}{\partial u^i} \bigg\vert_{u=\phi(p)} \end{align*}\]

so we have once again shown what we wanted.

Second definition using charts

The previous theorem proved that we can consider an alternate definition of the tangent space: $T_p M$ is the set of all linear maps $v : C^\infty(M) \to \R$ of the form

\[v(f) = \sum_{i=1}^m a^i \frac{\partial (f \circ \inv\phi)}{\partial u^i} \bigg\vert_{\u=\phi(p)} = \a \cdot \nabla (f \circ \inv \phi)\vert_{\phi(p)}\]

for some $\a = (a^1, …, a^m) \in \R^m$.

Note that this implies that $T_p M$ is an $m$-dimensional vector space, as it is spanned by the elements

\[\frac{\partial}{\partial x^1} \bigg\vert_{p}, ..., \frac{\partial}{\partial x^m} \bigg\vert_{p}\]

Product rule for tangent vectors

Let $v \in T_p M$ be a tangent vector to $M$ at $p$. Then

\[v(fg) = f(p) v(g) + v(f) g(p)\]

Suppose $\gamma \in C^\infty(J, M)$ is such that

\[v(h) = \frac{d}{dt}\bigg\vert_{t=0} h(\gamma(t))\]

then

\[\begin{align*} v(fg) &= \frac{d}{dt}\bigg\vert_{t=0} (fg) \circ \gamma(t) \\ &= \frac{d}{dt}\bigg\vert_{t=0} (f \circ \gamma(t))(g \circ \gamma(t)) \\ &= \frac{d}{dt}\bigg\vert_{t=0} (f \circ \gamma(t)) g(p) + \frac{d}{dt}\bigg\vert_{t=0} f(p)(g \circ \gamma(t)) \tag{product rule} \\ &= v(f) g(p) + f(p) v(g) \end{align*}\]

As a result, if $c$ is a constant function then $v(c) = 0$. This is because $v$ is linear, so

\[\begin{align*} cv(f) &= v(c \cdot f) \\ cv(f) &= v(c) f(p) + c(p) v(f) \\ cv(f) &= v(c) f(p) + cv(f) \\ 0 &= v(c) f(p) \end{align*}\]

this is true for all $f$, no matter what $f(p)$ is, so $v(c)$ must be $0$.

Another result is, if $f(p) = g(p) = 0$ then $v(fg) = 0$ regardless of $v(f)$ and $v(g)$.

Third definition using product rule

A linear map $v : C^\infty(M) \to \R$ is an element of $T_p M$ if and only if it satisfies the product rule at $p$, i.e.

\[v(fg) = f(p) v(g) + v(f) g(p)\]

We need a number of lemmas to prove that this is an equivalent characterization. For the following lemmas, assume that $v : C^\infty(M) \to \R$ is a linear map that satisfies the product rule at $p$ outlined above.

Hadamard’s lemma: First order approximation on manifolds

Let $U$ be an open ball around the origin in $\R^m$ and assume that $h \in C^\infty(U)$. Then there exist functions $h_1, …, h_m \in C^\infty(U)$ such that

\[h(\u) = h(\bzero) + \sum_{i=1}^m u^i h_i(\u)\]

and these functions satisfy

\[h_i(\bzero) = \frac{\partial h}{\partial u^i}(\bzero)\]

Proof.

We can parametrize the line from $\bzero$ to $\u = (u^1, …, u^m)$ in $\R^m$ by considering $su$ for $s \in [0, 1]$. Using this, we can apply the fundamental theorem of calculus to see that

\[\begin{align*} h(\u) - h(\bzero) &= \int_0^1 \frac{d}{dt}h(s\u) ds \\ &= \int_0^1 \u \cdot \nabla h(s\u) ds \tag{chain rule} \\ &= \sum_{i=1}^m u^i \int_0^1 \frac{\partial h}{\partial u^i}(s\u)ds \end{align*}\]

If we define each of our functions as $\ds h_i(\u) = \int_0^1 \frac{\partial h}{\partial u^i}(s\u)ds$, then

\[\begin{align*} h(\u) - h(\bzero) &= \sum_{i=1}^m u^i h_i(\u) \end{align*}\]

and

\[h_i(\bzero) = \int_0^1 \frac{\partial h}{\partial u^i}(s\bzero)ds = \int_0^1 \frac{\partial h}{\partial u^i}(\bzero) ds = \frac{\partial h}{\partial u^i}(\bzero)\]

as required.

Lemma: If $f_1 = f_2$ on a neighbourhood $U$ of $p$ then $v(f_1) = v(f_2)$

Let $f_1, f_2 \in C^\infty(M)$ be functions so that $f_1(u) = f_2(u)$ for all $u \in U$, where $U$ is some neighbourhood of $p$. Let $g = f_1 - f_2$, then $g$ is 0 on $U$. Let $\chi \in C^\infty(M)$ be a smooth function so that $\chi(p) = 1$ and $\chi(u) = 0$ for all $u \in M \setminus U$, i.e. it is a bump function for $p$ in $U$. Then $g$ is 0 on $U$ and $\chi$ is 0 on $M \setminus U$, so $g\chi$ is 0 on all of $M$, so $v(g\chi) = 0$. Then, using the product rule,

\[\begin{align*} 0 &= v(g\chi) \\ &= g(p) v(\chi) + v(g) \chi(p) \\ &= 0 \cdot v(\chi) + v(g) \cdot 1 \\ &= v(g) \\ &= v(f_1 - f_2) \\ &= v(f_1) - v(f_2) \tag{$v$ is linear} \end{align*}\]

so $v(f_1) = v(f_2)$.

This shows that $v$ can equivalently be defined as a function from $C^\infty(U)$ to $\R$ rather than from all of $C^\infty(M)$ to $\R$, since all that matters is how the input function behaves in a neighbourhood around $p$.

Lemma: We can lift $v$ to a map $\widetilde v$ that acts on Euclidean functions which satisfies the product rule

Let $(U, \phi)$ be a chart around $p$, and let $\widetilde U = \phi(U) \subseteq \R^m$ and $\widetilde p = \phi(p)$.

Suppose $\widetilde h \in C^\infty(\widetilde U)$ then we can define the function

\[\begin{align*} \widetilde v : C^\infty(\widetilde U) &\to \R \\ \widetilde h &\mapsto v(\widetilde h \circ \phi) \end{align*}\]

(Even though $\widetilde h \circ \phi$ is in $C^\infty(U)$ rather than $C^\infty(M)$, this function is still well-defined due to the previous lemma.)

Then if $\widetilde f, \widetilde g \in C^\infty(\widetilde U)$, then

\[\begin{align*} \widetilde v(\widetilde f \widetilde g) &= v((\widetilde f \widetilde g) \circ \phi) \\ &= v((\widetilde f \circ \phi)(\widetilde g \circ \phi)) \\ &= (\widetilde f \circ \phi(p)) v(\widetilde g \circ \phi)) + v(\widetilde f \circ \phi)) (\widetilde g \circ \phi(p)) \\ &= \widetilde f(\widetilde p) \widetilde v(\widetilde g) + \widetilde v(\widetilde f) \widetilde g(\widetilde p) \end{align*}\]

so $\widetilde v$ satisfies the product rule.

Proof of third definition

We know that tangent vectors by our previous definition satisfy the product rule, so we know one direction of the equivalence.

For the other direction, let $(U, \phi)$ be a chart around $p$ and let $\widetilde U = \phi(U)$ and $\widetilde p = \phi(p)$. Assume without loss of generality that $\widetilde p = \bzero$.

Define $\widetilde v$ using the previous lemma. Then if $f \in C^\infty(M)$, let $\widetilde f = f \circ \inv\phi$ be its lift to Euclidean space. By Hadamard’s lemma, we know that there exist functions $f_1, …, f_m \in C^\infty(\widetilde U)$ so that

\[\widetilde f(\u) = \widetilde f(\bzero) + \sum_{i=1}^m u^i f_i(\u) \text{ and } f_i(\bzero) = \frac{\partial \widetilde f}{\partial u^i}(\bzero)\]

Since $\widetilde v$ satisfies the product rule, we also know that $\widetilde v(c) = 0$ for any constant function $c$. Therefore,

\[\begin{align*} v(f) &= v(f \circ \inv \phi \circ \phi) \\ &= \widetilde v(\widetilde f) \\ &= \widetilde v\left(\widetilde f(\bzero) + \sum_{i=1}^m u^i f_i\right) \\ &= \widetilde v(\widetilde f(\bzero)) + \sum_{i=1}^m \widetilde v(u^i f_i) \tag{$\widetilde v$ is linear} \\ &= \sum_{i=1}^m \widetilde v(u^i f_i) \tag{$\widetilde v(c) = 0$ for constant $c$} \\ &= \sum_{i=1}^m (u^i(\bzero) \widetilde v(f_i) + \widetilde v(u^i) f_i(\bzero)) \tag{product rule} \\ &= \sum_{i=1}^m (0 \cdot \widetilde v(f_i) + \widetilde v(u^i) \frac{\partial \widetilde f}{\partial u^i}) \\ &= \sum_{i=1}^m \widetilde v(u^i) \frac{\partial (f \circ \inv\phi)}{\partial u^i} \end{align*}\]

So if $a^i = \widetilde v(u^i)$, then

\[v(f) = \sum_{i=1}^m a^i \frac{\partial (f \circ \inv\phi)}{\partial u^i}\]

which is exactly the form of a tangent vector, by the second definition.

Tangent map

Let $F \in C^\infty(M, N)$. For any $p \in M$, we define the tangent map to be the linear map

\[\begin{align*} T_p F : T_p M &\to T_{F(p)} N \\ v(g) &\mapsto v(g \circ F) \end{align*}\]

We can verify that this definition sends tangent vectors to tangent vectors:

Let $v \in T_p M$ and let $w = T_p F(v) \in T_{F(p)} N$.

Let $g, h \in C^\infty(N)$ and $a, b \in \R$. Then,

\[\begin{align*} w(ag + bh) &= v(ag \circ F + bh \circ F) \\ &= v(ag \circ F) + v(bh \circ F) \\ &= aw(g) + bw(h) \end{align*}\]

so $w$ is linear, and

\[\begin{align*} w(gh) &= v((gh) \circ F) \\ &= v((g \circ F)(h \circ F)) \\ &= (g \circ F(p)) v(h \circ F) + v(g \circ F) (h \circ F(p)) \\ &= g(F(p)) w(h) + w(g) h(F(p)) \end{align*}\]

so $w$ satisfies the product rule at $F(p)$. Thus, $w \in T_{F(p)} N$.

Tangent map and curves

Suppose $v \in T_p M$ is represented by a curve $\gamma : J \to M$ where $J \subseteq \R$ is an open interval around $0$. Then $(T_p F) (v)$ is represented by the curve $F \circ \gamma : J \to N$

Suppose $\ds v(g) = \frac{d}{dt}\bigg\vert_{t=0} g(\gamma(t))$, then

\[\begin{align*} (T_p F)(v)(g) &= v(g \circ F) = \frac{d}{dt}\bigg\vert_{t=0} (g \circ F)(\gamma(t)) \\ &= \frac{d}{dt}\bigg\vert_{t=0} g(F \circ \gamma(t)) \end{align*}\]

which is what we wanted to show.

Push-forwards and pull-backs

Let $F \in C^\infty(M, N)$. We can use $F$ to transport between certain “objects defined on $M$” and “objects defined on $N$”:

Then, using this language, if $v \in T_p M$ then the push-forward of $v$ by $F$ is $F_\ast v = (T_p F)(v) \in T_{F(p)} N$.

Note that $F_\ast v(g) = v(F^\ast g)$

Chain rule for tangent map

Let $M, N, Q$ be manifolds and assume that $F \in C^\infty(M, N)$ and $G \in C^\infty(N, Q)$. Then if $p \in M$, then

\[T_p (G \circ F) = T_{F(p)} G \circ T_p F\]

This is because if $v \in T_p M$ and $g \in C^\infty(Q)$, then

\[\begin{align*} (T_p (G \circ F))(v)(g) &= v((g \circ G) \circ F) \\ &= (T_p F)(v)(g \circ G) \\ &= (T_{F(p)} G)((T_p F)(v))(g) \\ &= (T_{F(p)} G \circ T_p F)(v)(g) \end{align*}\]

Properties of the tangent map

The tangent map of the identity map $\id_M : M \to M$ is the identity map on the tangent space:

\[T_p \id_M = \id_{T_p M}\]

This is because

\[(T_p \id_M)(v)(g) = v(g \circ \id_M) = v(g) = \id_{T_p M} (v)(g)\]

If $F \in C^\infty(M, N)$ is a diffeomorphism then $T_p F$ is a linear isomorphism between the vector spaces $T_p M$ and $T_{F(p)} $, and $(T_p F)^{-1} = T_{F(p)}(\inv F)$. We use the previous fact to prove this:

\[\begin{align*} \id_M &= \inv F \circ F \\ T_p \id_M &= T_p (\inv F \circ F) \\ \id_{T_p M} &= T_{F(p)} \inv F \circ T_p F \end{align*}\]

A similar derivation holds for $F \circ \inv F$ as well, so $T_{F(p)} \inv F$ is the inverse of $T_p F$. Since $T_p F$ is invertible, it is an isomorphism.

Suppose $q \in N$ and let $F : M \to N$ be the constant map so that for all $p \in M$, $F(p) = q$. Then $T_p F$ is the zero map for all $p \in M$. To prove this, we use the fact that tangent vectors vanish for constant functions:

\[T_p F(v)(g) = v(\underbrace{g \circ F}_{\text{constant}}) = 0\]

Tangent map and Jacobian matrix

Suppose $U \subseteq \R^m$ and $V \subseteq \R^n$ are open, let $F \in C^\infty(U, V)$, and let $p \in U$. Then,

\[T_p F \left( \frac{\partial}{\partial x^i}\bigg\vert_p \right) = \sum_j \frac{\partial y^j}{\partial x^i}\bigg\vert_p \frac{\partial}{\partial y^j}\bigg\vert_{F(p)}\]

where $y = F(x)$ and $y^j = F^j(x)$.

We proved earlier that

\[\frac{\partial}{\partial x^1}\bigg\vert_p, ..., \frac{\partial}{\partial x^m}\bigg\vert_p\]

form a basis for $T_p U$ Thus, this theorem shows that in this basis, the matrix for $T_p F$ is the Jacobian matrix. Therefore $T_p F = D_p F$ for functions between Euclidean spaces.

Proof.

Suppose $g \in C^\infty(N)$. Then,

\[\begin{align*} T_p F \left( \frac{\partial}{\partial x^i}\bigg\vert_p \right)(g) &= \frac{\partial}{\partial x^i}\bigg\vert_p (g \circ F) \\ &= \frac{\partial}{\partial x^i}\bigg\vert_p (g \circ y) \\ &= \sum_j \frac{\partial y^j}{\partial x^i}\bigg\vert_p \frac{\partial g}{\partial y^j}\bigg\vert_{F(p)} \end{align*}\]

Redefinitions in terms of the tangent map

Now that we see the analogy between the tangent map and the Jacobian matrix, we can make many redefinitions:

Suppose $F \in C^\infty(M, N)$, then

Tangent spaces and submanifolds

Let $S \subseteq M$ be a submanifold and let $p \in S$. Let $i : S \to M$ be the inclusion map, so for $x \in S$, $i(x) = x \in M$. Then we identify the tangent space $T_p S$ of $S$ with the image $(T_p i)(T_p S) \subseteq T_p M$.

Regular points are in the kernel of the tangent map

Suppose $F \in C^\infty(M, N)$ where $\dim M = m$ and $\dim N = n$. Let $q \in N$ be a regular value of $F$, and let $S = \inv F\curlies{q}$ (so $S$ is a submanifold).

Then for every $p \in S$,

\[T_p S = \ker(T_p F) \subseteq T_p M\]

Proof.

$q$ is a regular value, so $T_p F$ is surjective for every $p \in \inv F \curlies q$. Thus, $\rank T_p F = n$, so $\ker(T_p F)$ has dimension $m - n$ (rank-nullity). By the regular value theorem, $\dim S = \dim \ker(T_p F)$. We also know that the tangent space of a manifold has the same dimension as the manifold, so $\dim T_p S = \dim S = \dim \ker(T_p F) = m - n$.

If we show that $T_p S \subseteq \ker(T_p F)$, then we will be done, because a vector subspace of $\ker(T_p F)$ with the same dimension must be all of $T_p F$.

Using the chain rule, we know that \(T_p F \vert_{T_p S} = T_p F \circ T_p i = T_{i(p)} F \circ T_p i = T_p (F \circ i)\) $F \circ i$ is constant on $S$ since $S = \inv F \curlies q$. The tangent map of a constant function is 0 so $T_p (F \circ i) = 0$ on all of $T_p S$. However, the inclusion map has trivial null space, so this means that $T_p F$ is zero on $T_p S$, so $T_p S \subseteq \ker(T_p F)$.

Corollary

Suppose $V \subseteq \R^\ell$ is open and $F \in C^\infty(V, \R^k)$. Let $q \in \R^k$ be a regular value of $F$ and let $M = \inv F \curlies q \subseteq \R^\ell$, so $M$ is a submanifold. If $p \in M$, then \(T_p M = \ker(D_p F) \subseteq T_p \R^\ell = \R^\ell\)

Example: Tangent space to $S^n$

Let $p \in S^n \subseteq \R^{n+1}$. We know that if we consider the squared norm,

\[N(x^1, ..., x^{n+1}) = (x^1)^2 + ... + (x^{n+1})^2\]

then $S = \inv N \curlies 1$. Then the Jacobian of $N$ at $p$ is

\[\begin{align*} D_p F(x) &= \lim_{t \to 0} F(p + tx) \\ &= 2p \cdot x \end{align*}\]

so the tangent space to $S^n$ at $p$ is

\[T_p S^n = \ker(D_p F) = \curlies{x \in \R^{n+1} : p \cdot x = 0}\]

i.e. the set of all vectors orthogonal to $p$.

Critical points of a function on a submanifold

Suppose $S$ is a submanifold of $M$ with inclusion map $i : S \to M$, and $h \in C^\infty(M)$ is a smooth function. Let $f = h \vert_S = h \circ i$.

Define the set of critical points:

\[\Crit(f) = \curlies{p \in S : \rank_p f < 1}\]

The rank of $f$ is less than one if and only if $T_p F$ is the zero map. Also, by the chain rule, $T_p f = T_p h \circ T_p i$. Then we see that

\[\Crit(f) = \curlies{p \in S : \text{image}(T_p i) = T_p S \subseteq \ker(T_p h)}\]

For example, let $S \subseteq \R^3$ be a submanifold and let $h(x, y, z) = z$. We know that $T_p h \cong D_p h = \begin{pmatrix} 0 & 0 & 1 \end{pmatrix}$, so

\[\ker(T_p h) \cong \curlies{(v_1, v_2, v_3) : v_3 = 0} = \text{the } xy \text{-plane}\]

So the critical points of $f$ are

\[\Crit(f) = \curlies{p \in S : T_p S \subseteq \text{the } xy \text{-plane}}\]

Matrix Lie groups

A matrix Lie group is a submanifold $G \subseteq \Mat_\R(n)$ that is also a subgroup with respect to multiplication, i.e.

Examples:

Matrix Lie algebras

If $G$ is a matrix Lie group, then the Lie algebra of $G$ is $T_I G \subseteq \Mat_\R(n)$, denoted as $\mathfrak g$. (Similarly, \mathfrak is used for other symbols denoting Lie algebras.)

Examples:

$SL(n, \R)$ is a matrix Lie group

To show that $SL(n, \R)$ is a submanifold, we will show that $1$ is a regular value of the determinant function.

\[\begin{align*} D_A \det(X) &= \frac{d}{dt}\bigg\vert_{t=0} \det(A + tX) \end{align*}\]

The determinant is the product of the roots of the characteristic polynomial, which we can find by solving $\det(A + tX - \lambda I) = 0$.

Let’s consider $A = I$. Then we want to solve

\[\begin{align*} \det(I + tX - \lambda I) &= 0 \\ t^n\det\left(X - \frac{\lambda - 1}{t} I\right) &= 0 \end{align*}\]

i.e., $\lambda$ is an eigenvalue of $I + tX$ if and only if $\ds \frac{\lambda - 1}{t}$ is an eigenvalue of $X$. So the eigenvalues of $I + tX$ are

\[1 + t\mu_1, ..., 1 + t\mu_k\]

where $\mu_1, …, \mu_k$ are the eigenvalues of $X$. So considering the determinant,

\[\begin{align*} \det(I + tX) &= \prod_{i=1}^k (1 + t\mu_i) \\ \frac{d}{dt}\bigg\vert_{t=0} \det(I + tX) &= \sum_{i=1}^k \mu_i \\ (D_I \det)(X)&= \text{tr} X \end{align*}\]

Now considering the derivative at any arbitrary matrix $A \in SL(n, \R)$,

\[\begin{align*} (D_A \det)(X) &= \frac{d}{dt}\bigg\vert_{t=0} \det(A + tX) \\ &= \frac{d}{dt}\bigg\vert_{t=0} \det(A(I + t\inv A X)) \\ &= \det(A) \cdot \frac{d}{dt}\bigg\vert_{t=0} \det(I + t\inv A X) \\ &= \det(A) \cdot (D_I \det)(\inv A X) \\ &= \det(A) \text{tr}(\inv A X) \end{align*}\]

Suppose $\det(A) = 1$, then

\[(D_A \det)(X) = \text{tr}(X)\]

which is surjective onto $\R$, so $\det$ has maximal rank on $SL(n, \R) = \inv \det \curlies 1$. Thus, $1$ is a regular value, so $SL(n, \R)$ is a submanifold.

Calculating its Lie algebra,

\[\begin{align*} \mathfrak{sl}(n, \R) &= T_I SL(n, \R) \\ &= \ker(D_I \det) \\ &= \curlies{X : \text{tr}(X) = 0} \end{align*}\]

Properties of matrix Lie algebras

Let $G$ be a matrix Lie group and let $\mathfrak g = T_I G$ be its Lie algebra. Then

  1. if $A \in G$, $T_A G = \curlies{AX : X \in \mathfrak g} = \curlies{XA : X \in \mathfrak g}$
  2. if $A \in G$ and $X \in \mathfrak g$, then $AX\inv A \in \mathfrak g$
  3. if $X, Y \in \mathfrak g$ then so is their commutator, $[X, Y] = XY - YX \in \mathfrak g$

Proof.

Part 1

Let $Y \in T_A G$, then there is some path $\gamma : (-\delta, \delta) \to G$ with $\gamma(0) = A$ that represents $Y$. Then, since $G$ is a Euclidean space, we can identify $Y$ with

\[Y \cong \frac{d}{dt}\bigg\vert_{t=0} \gamma(t)\]

Define the path $\lambda(t) = \inv A \gamma(t)$, then $\lambda$ is also a path in $G$ and $\lambda(0) = I$. Then we can identify define the tangent vector $X \in \mathfrak g$ represented by $\lambda$, and we can identify $X$ with

\[X \cong \frac{d}{dt}\bigg\vert_{t=0} \lambda(t) = \frac{d}{dt}\bigg\vert_{t=0} \inv A \gamma(t) = \inv A \frac{d}{dt}\bigg\vert_{t=0} \cong \inv A Y\]

so $Y = AX$.

The other identity can be proven similarly if we instead set $\lambda(t) = \gamma(t) \inv A$.

Part 2

Suppose $A \in G$ and $X \in \mathfrak g$. Then $AX \in T_A G = \curlies{YA : Y \in \mathfrak g}$, so $AX = YA$ for some $Y \in \mathfrak g$. Then $AX\inv A = Y \in \mathfrak g$.

Part 3

Suppose $X, Y \in \mathfrak g$. Then there is some curve $\gamma : (-\delta, \delta) \to G$ with $\gamma(0) = I$ that represents $X$, so we can identify

\[X \cong \frac{d}{dt}\bigg\vert_{t=0} \gamma(t)\]

We can define a new curve

\[\begin{align*} \lambda : (-\delta, \delta) &\to \mathfrak g \\ t &\mapsto \gamma(t) Y \inv{\gamma(t)} \end{align*}\]

Since $\gamma(t) \in G$, by part 2 we know that $\lambda(t) = \gamma(t) Y \inv{\gamma(t)} \in \mathfrak g$.

We can find the derivative of $\inv{\gamma(t)}$:

\[\begin{align*} (\gamma \inv{\gamma}) &= I \\ D_t (\gamma \inv{\gamma}) &= D_t I \\ (D_t \gamma) \inv{\gamma} + \gamma (D_t \inv{\gamma}) &= 0 \\ \gamma (D_t \inv{\gamma}) &= -(D_t \gamma) \inv{\gamma} \\ D_t \inv{\gamma} &= -\inv{\gamma} (D_t \gamma) \inv{\gamma} \\ \end{align*}\]

Using this, we can find the derivative of $\lambda$:

\[\begin{align*} D_t \lambda &= D_t (\gamma Y \inv{\gamma}) \\ &= (D_t \gamma) \cdot Y \inv{\gamma} + \gamma \cdot (D_t X) \inv{\gamma} + \gamma Y \cdot (D_t \inv{\gamma}) \\ &= (D_t \gamma) \cdot Y \inv{\gamma} + \gamma Y \cdot (-\inv{\gamma} (D_t \gamma) \inv{\gamma} )\\ D_0 \lambda &= D_0 \gamma \cdot Y \cdot \inv{\gamma(0)} - \gamma(0) \cdot Y \cdot \inv{\gamma(0)} \cdot D_0 \gamma \cdot \inv{\gamma(0)} \\ &= X Y \inv I - I Y \inv I X \inv I \\ &= XY - YX \end{align*}\]

The derivative of a curve in a vector space is still in that vector space, so $[X, Y] = XY-YX = D_0 \lambda \in \mathfrak g$.