= Solution
The real <vector space> of <symmetric matrices> has dimension $n(n+1)/2$. On its invertible part the derivative of the <determinant> is
$$
D(\det)_B(C)=\det(B)\operatorname{tr}(B^{-1}C).
$$
At a determinant-one symmetric matrix, this derivative is nonzero: the symmetric direction $C=B$ gives value $n$. The <regular level set theorem> therefore yields
$$
\boxed{\dim M=\frac{n(n+1)}2-1,\qquad
T_BM=\{C=C^T:\operatorname{tr}(B^{-1}C)=0\}.}
$$
This applies to every signature component, not only positive-definite matrices.
The <special linear congruence action on symmetric matrices> preserves symmetry because $(ABA^T)^T=ABA^T$, and preserves determinant because $\det(ABA^T)=(\det A)^2\det B=1$. The identity acts trivially and
$$
A_1\cdot(A_2\cdot B)=A_1A_2B(A_1A_2)^T=(A_1A_2)\cdot B.
$$
These verify a smooth left <Lie group action> of the <special linear group>. By <Sylvester's law of inertia>, it also preserves the numbers of positive and negative eigenvalues.
For $n=2$, parametrize the <symmetric matrices> by
$$
B=\begin{pmatrix}t+x&y\\y&t-x\end{pmatrix},\qquad
\det B=t^2-x^2-y^2.
$$
Thus $M$ is exactly the <two-sheeted hyperboloid>
$$
\boxed{t^2-x^2-y^2=1,\qquad t>0\text{ or }t<0.}
$$
The two sheets consist respectively of positive-definite and negative-definite matrices. Each is preserved by the <special linear congruence action on symmetric matrices>. They are individually transitive: on the positive sheet $B=AA^T$ with $A=B^{1/2}$ and $\det A=1$; on the negative sheet use $B=-AA^T$. The stabilizer of either $I$ or $-I$ is $SO(2)$.
Differentiating the left <Lie group action> along $\exp(sX)$ gives the <fundamental vector field>
$$
u_X(B)=XB+BX^T.
$$
For the three matrices in the question, their coordinate components are
$$
\begin{aligned}
u_1&=y\partial_t+y\partial_x+(t-x)\partial_y,\\
u_2&=2x\partial_t+2t\partial_x,\\
u_3&=y\partial_t-y\partial_x+(t+x)\partial_y.
\end{aligned}
$$
A sign convention matters here. With the usual <Lie bracket of vector fields> $[U,V]f=U(Vf)-V(Uf)$, the fundamental fields of this left action obey $[u_X,u_Y]=-u_{[X,Y]}$. Indeed, for linear coordinate fields $u_X(z)=R_Xz$, the bracket has coefficient $(R_YR_X-R_XR_Y)z$. To obtain a <Lie algebra representation> rather than an anti-representation, use the <infinitesimal left-action sign convention>
$$
v_X(B)=\left.\frac{d}{ds}\right|_{s=0}e^{-sX}Be^{-sX^T}=-u_X(B).
$$
An explicit answer is therefore
$$
\boxed{\begin{aligned}
v_1&=-y\partial_t-y\partial_x+(x-t)\partial_y,\\
v_2&=-2x\partial_t-2t\partial_x,\\
v_3&=-y\partial_t+y\partial_x-(t+x)\partial_y.
\end{aligned}}
$$
These are tangent to the <two-sheeted hyperboloid>: applying each to $Q=t^2-x^2-y^2$ gives zero. For example,
$$
v_1(Q)=-2ty+2xy-2y(x-t)=0.
$$
They can be written on either sheet using $t=\pm\sqrt{1+x^2+y^2}$:
$$
v_1=-y\partial_x+(x-t)\partial_y,\qquad
v_2=-2t\partial_x,\qquad
v_3=y\partial_x-(t+x)\partial_y.
$$
Here the omitted $\partial_t$ component is determined by tangency, and derivatives of $t(x,y)$ must be included when computing brackets in these coordinates.
Direct matrix multiplication gives
$$
[\tau_2,\tau_1]=2\tau_1,\qquad
[\tau_2,\tau_3]=-2\tau_3,\qquad
[\tau_1,\tau_3]=\tau_2.
$$
Direct differentiation of the displayed <vector fields> gives exactly
$$
\boxed{[v_2,v_1]=2v_1,\qquad [v_2,v_3]=-2v_3,\qquad [v_1,v_3]=v_2.}
$$
For instance, in ambient $(t,x,y)$ coordinates, $[v_1,v_3]=(-2x,-2t,0)=v_2$. These are the defining relations of the <sl2 Lie algebra>. The three fields are linearly independent over constant real coefficients: if $a v_1+b v_2+c v_3$ vanishes on a sheet, its $\partial_x$ coefficient $(-a+c)y-2bt$ forces $b=0,c=a$, and its $\partial_y$ coefficient then equals $-2at$, forcing $a=c=0$. Thus the representation is \b[faithful]. Their pointwise span need only have dimension two, consistent with the dimension of each sheet.
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