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At $p$, choose local normal coordinates with $g_{\alpha \beta }(p)=\mathrm{diag}(-1,1)$. Let $e_0,e_1$ be their coordinate basis. The vectors $V=e_0+e_1$ and $W=e_0-e_1$ have zero metric tensor square, and their component matrix has determinant $-2\ne 0$. Thus two independent null vectors form a basis of the tangent space. Normal coordinates are only needed at the chosen point, not on a whole neighbourhood with flat metric.

For $p\in \mathrm{Fix}(\sigma )$, differentiation gives $(d{\sigma }_p{)}^2=I$ and $(d{\sigma }_p{)}^{\ast }{\omega }_p=-{\omega }_p$. Thus $d{\sigma }_p$ is an anti-symplectic involution of the symplectic vector space $T_{p}M$. By part (a), its $+1$ eigenspace is a Lagrangian subspace.

In the local manifold chart supplied in the question, $\sigma$ is linear, so its fixed-point set is the intersection of the chart with this $+1$ eigenspace. Therefore the fixed-point set is an embedded submanifold andEvery such tangent space has half the ambient dimension and the symplectic form vanishes on it. Hence every nonempty fixed-point set is a Lagrangian submanifold. The fixed-point set itself need not be connected.

Use ${\mathbb{C}}^2$ with the standard symplectic form ${\omega }_0=d{x}_1\wedge d{y}_1+d{x}_2\wedge d{y}_2$ and the compatible almost complex structure given by multiplication by $i$. Consider the embedded submanifoldIn coordinates $(x_1,y_1,x_2,y_2)$, its tangent space is $W=\{(a,b,b,0):a,b\in \mathbb{R}\}$. Applying $J$ gives $J(a,b,b,0)=(-b,a,0,b)$. If this is again in $W$, its fourth coordinate forces $b=0$, and its second and third coordinates then force $a=0$. Thus $W\cap JW=0$ and ${\dim }_{\mathbb{R}}L=2$: $L$ is a totally real submanifold.

However, the pullback of a differential form under the parametrization isThus this totally real submanifold is not a Lagrangian submanifold.

For an embedded submanifold $X\subseteq {\mathbb{R}}^N$ and $p\in X$, the tangent space isEquivalently, in a local parametrization $\phi :U\subseteq {\mathbb{R}}^d\to X$ with $\phi (u)=p$, one has $T_{p}X=\mathrm{im}D{\phi }_u$.

Let $N=(0,\dots ,0,1)$ and $S=(0,\dots ,0,-1)$ be the poles of $S^n$. The open sets $U_N=S^n\setminus \{N\}$ and $U_S=S^n\setminus \{S\}$ cover the sphere. Stereographic projection gives coordinate chartsfrom these sets to ${\mathbb{R}}^n$. Their inverses areOn the overlap, the transition map iswhich is a smooth diffeomorphism of ${\mathbb{R}}^n\setminus \{0\}$. These two compatible charts make $S^n$ a smooth manifold of dimension $n$.

Now let $G$ be an $m$-dimensional Lie group and choose a basis $e_1,\dots ,e_m$ of its tangent space $T_{e}G$ at the identity. Definewhere $L_g$ is Left translation on a Lie group. Smoothness of multiplication makes each $E_i$ a smooth left-invariant vector field, and invertibility of $(d{L}_g{)}_e$ makes $E_1(g),\dots ,E_m(g)$ a basis of $T_{g}G$ at every point. Thus the $E_i$ form a global frame and every Lie group is a parallelizable manifold.

The columns of a matrix in the special unitary group $SU(2)$ are orthonormal and its determinant is one. Consequently every element has the unique formThe pair $(z_1,z_2)\in {\mathbb{C}}^2\cong {\mathbb{R}}^4$ therefore identifies $SU(2)$ diffeomorphically with $S^3$. The Lie-group construction then proves that $S^3$ is parallelizable; this is the SU(2) as the three-sphere identification.

Another example is $S^1$, which is the Lie group $U(1)$. Explicitly, at $(x,y)\in S^1$ the vectoris smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another parallelizable sphere.

For each $p\in S$, the first fundamental form is defined without coordinates bythe restriction of the Euclidean inner product to the tangent space of $S$.

If $\varphi (u,v)$ is a local parametrization andthenTaking their dot product giveswhere

For an affine algebraic group $G$, its Lie algebra is the tangent space at the identity,Equivalently, it is the space of left-invariant derivations of the coordinate ring $\mathbb{C}[G]$. The Lie bracket is the commutator of derivations,It is antisymmetric because $[Y,X]=-[X,Y]$. Associativity of composition givesafter all six triple products cancel in pairs, proving the Jacobi identity. This construction is the Lie algebra of an affine algebraic group.

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

A surjective submersion between Riemannian manifolds is Riemannian when its derivative is a linear isometry from the orthogonal complement of its kernel onto the target tangent space. Its vertical space is the kernel and its horizontal space is the orthogonal complement.

For an embedded surface with unit normal vector $\mathbf{n}$, the surface gradient is ${\nabla }_{s}f=(\mathbf{I}-\mathbf{n}{\mathbf{n}}^T)\nabla f$. It projects the ambient gradient onto the tangent space and is independent of how $f$ is extended normally off the surface.

Initial position ${\varphi }_0$ for a wave map takes values in the target, and initial velocity ${\varphi }_1$ belongs to its tangent space at ${\varphi }_0(x)$. For $S^2$ this means $|{\varphi }_0|=1$ and ${\varphi }_0\cdot {\varphi }_1=0$. Localized data agree with a constant map and zero velocity outside a compact set; the map itself need not be a compactly supported ambient function.