Solution
= Solution
A smooth <vector field> on a <smooth manifold> $M$ is a smooth section of the <tangent bundle>: it assigns $v(p)\in T_pM$ to each $p$, smoothly in local coordinates. In a coordinate chart it has the form $v=\sum_i a_i\partial/\partial x_i$, with smooth coefficients $a_i$. Equivalently, it acts on smooth functions as a derivation, $v(fh)=v(f)h+fv(h)$.
For a <Lie group> $G$, write $L_a(h)=ah$ for <Left translation on a Lie group>. A <left-invariant vector field> satisfies
$$
(dL_a)_h v(h)=v(ah)\qquad(a,h\in G).
$$
Thus it is determined by its value at the identity. Given the printed $X\in T_gG$, first translate it to the identity:
$$
\xi=(dL_{g^{-1}})_gX\in T_eG.
$$
The unique <left-invariant vector field> with value $X$ at $g$ is
$$
\boxed{v_X(h)=(dL_h)_e\xi=(dL_{hg^{-1}})_gX.}
$$
The smoothness of multiplication makes this a smooth <vector field>, the chain rule proves left invariance, and setting $h=g$ gives $v_X(g)=X$.
We prove the <completeness of left-invariant vector fields>: this <left-invariant vector field> is a <complete vector field>. The <Picard-Lindelöf theorem>, applied in a local chart, gives a unique local <integral curve of a vector field> $u:(-\varepsilon,\varepsilon)\to G$ through $e$. For every $h\in G$, the curve $t\mapsto hu(t)$ is an <integral curve of a vector field> through $h$, because
$$
\frac{d}{dt}(hu(t))=(dL_h)_{u(t)}v_X(u(t))=v_X(hu(t)).
$$
Crucially, the same interval $(-\varepsilon,\varepsilon)$ works for every initial point.
Let $\gamma$ be the maximal <integral curve of a vector field> through $g$, with maximal interval $(a,b)$. If $b<\infty$, choose $t_0\in(a,b)$ with $b-t_0<\varepsilon/2$. The curve
$$
t\longmapsto\gamma(t_0)u(t-t_0)
$$
exists on $(t_0-\varepsilon,t_0+\varepsilon)$ and agrees with $\gamma$ on the overlap by uniqueness of the local <ordinary differential equation>. It extends $\gamma$ past $b$, a contradiction. The same argument at a finite $a$ excludes that possibility. Thus $(a,b)=\mathbb R$, and uniqueness on overlapping intervals gives uniqueness on all of $\mathbb R$.
After establishing completeness, uniqueness also gives the <one-parameter subgroup> law for the global curve through $e$:
$$
u(s+t)=u(s)u(t).
$$
Both sides, as curves in $t$, are <integral curves of a vector field> through $u(s)$ at $t=0$. Defining the <Exponential map of a Lie group> by $u(t)=\exp(t\xi)$, the answer is
$$
\boxed{\gamma_g(t)=g\exp(t\xi),\qquad \xi=(dL_{g^{-1}})_gX,\qquad t\in\mathbb R.}
$$
\b[The identity component $G^\circ$ is an open normal subgroup, and every open identity neighbourhood generates it.] Let $G^\circ$ be the <connected component> of $e$. The product of connected spaces is connected, so the image of $G^\circ\times G^\circ$ under multiplication is connected and contains $e$. It is therefore contained in $G^\circ$. Inversion has the same property. Thus $G^\circ$ is a <subgroup>.
A <smooth manifold> is locally connected. In particular, a coordinate neighbourhood of $e$ can be chosen homeomorphic to an open ball, so there is a connected open neighbourhood $O$ of $e$ contained in $G^\circ$. Its translates $hO$, $h\in G^\circ$, are open and lie in $G^\circ$, and cover $G^\circ$. Therefore the <identity component of a Lie group> is open in $G$. Conjugation by any $a\in G$ is a <homeomorphism> fixing $e$, so it maps $G^\circ$ into itself; conjugation by $a^{-1}$ gives the reverse inclusion. Hence $G^\circ$ is a <normal subgroup>.
If $U$ is an open neighbourhood of $e$ in $G^\circ$, let $H=\langle U\rangle$, allowing inverses in the meaning of generated <subgroup>. For each $h\in H$, $hU$ is open and lies in $H$, so $H=\bigcup_{h\in H}hU$ is open in $G^\circ$. Every other left <coset> of $H$ is also open. Thus $H$ is both open and closed in the connected space $G^\circ$. It is nonempty, so $H=G^\circ$.
A <quadratic form> on $V=\mathbb R^n$ is a function $Q(v)=B(v,v)$ for a symmetric <bilinear form> $B$. Equivalently, $Q(tv)=t^2Q(v)$ and the polarization
$$
B(u,v)=\frac{Q(u+v)-Q(u)-Q(v)}2
$$
is a <bilinear map>. In coordinates there is a unique real <symmetric matrix> $A$ with $Q(v)=v^TAv$. No assumption of nondegeneracy or positive definiteness is needed.
An element $g\in\operatorname{GL}_n(\mathbb R)$ stabilizes $Q$ when $Q(gv)=Q(v)$ for every $v$. By the <polarization identity>, this is equivalent to preserving $B$, or in coordinates to
$$
G_Q=\{g\in\operatorname{GL}_n(\mathbb R):g^TAg=A\}.
$$
This is a closed <Matrix Lie group>. The <Lie algebra of a quadratic-form stabilizer> is
$$
\boxed{T_eG_Q=\{X\in M_n(\mathbb R):X^TA+AX=0\}.}
$$
To prove necessity, differentiate $g(t)^TAg(t)=A$ along any smooth curve in $G_Q$ with $g(0)=I$, $g'(0)=X$. The derivative at zero is $X^TA+AX$.
To prove sufficiency, suppose $X^TA+AX=0$ and consider the <matrix exponential> $g(t)=e^{tX}$. Then
$$
\frac{d}{dt}\bigl(e^{tX^T}Ae^{tX}\bigr)=e^{tX^T}(X^TA+AX)e^{tX}=0.
$$
Its value at zero is $A$, so $e^{tX}\in G_Q$ for every real $t$. This curve has derivative $X$ at zero, proving the claimed <tangent space> description even for a degenerate <quadratic form>.
Equivalently, the condition is $B(Xu,v)+B(u,Xv)=0$ for all $u,v$. For a nondegenerate <quadratic form> it is the corresponding <Special orthogonal Lie algebra>. To see the degenerate case explicitly, choose a <basis> with
$$
A=\begin{pmatrix}D&0\\0&0\end{pmatrix},\qquad D=\operatorname{diag}(I_r,-I_s).
$$
Writing $X=\begin{pmatrix}P&R\\S&T\end{pmatrix}$, the condition becomes
$$
P^TD+DP=0,\qquad R=0,
$$
while $S,T$ are arbitrary. Thus the nullspace of the <quadratic form> is invariant, but arbitrary infinitesimal maps into it are allowed. When $A=0$, the formula correctly gives $G_Q=\operatorname{GL}_n(\mathbb R)$ and $T_eG_Q=M_n(\mathbb R)$.