Let $M$ be a differentiable manifold of dimension $n$. Consider an atlas $\mathcal{U}=\{U_\alpha\}_{\alpha\in\mathcal{A}}$ along with coordinates $x_\alpha^1,\cdots,x_\alpha^n$ in $U_\alpha$. For $x=(x_\alpha^1(x),\cdots,x_\alpha^n(x))\in U_\alpha$, a tangent vector is given by
$$v=\sum_{j=1}^nv_\alpha^j\frac{\partial}{\partial x_\alpha^j}.$$
If $x\in U_\alpha\cap U_\beta$, then $v$ is also written as
$$v=\sum_{j=1}^nv_\beta^j\frac{\partial}{\partial x_\beta^j}.$$
Here, the change of coordinates is given by
$$v_\beta^j=vx_\beta^j=\sum_{k=1}^nv_\alpha^k\frac{\partial x_\beta^j}{\partial x_\alpha^k}.$$
For $x\in U_\alpha\cap U_\beta$ and $f=(f^1,\cdots,f^n)\in\mathbb{R}^n$, define
\begin{align*}
h_{\alpha\beta}(x)(f)&=\left(\sum_{k=1}^n\frac{\partial x_\beta^1}{\partial x_\alpha^k}f^k,\cdots,(\sum_{k=1}^n\frac{\partial x_\beta^n}{\partial x_\alpha^k}f^k\right)\\
&=\begin{pmatrix}
\frac{\partial x_\beta^1}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^1}{\partial x_\alpha^n}\\
\vdots & \ddots & \vdots\\
\frac{\partial x_\beta^n}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^n}{\partial x_\alpha^n}
\end{pmatrix}\begin{pmatrix}
f^1\\
\vdots\\
f^n
\end{pmatrix}
\end{align*}
Hence $h_{\alpha\beta}:U_\alpha\cap U_\beta\longrightarrow\mathrm{Aut(\mathbb{R}^n)}$. The resulting bundle over $M$ with fibre $F=\mathbb{R}^n$ is called the tangent bundle of $M$ and is denoted by $TM$. Note that $TM$ is the set of all tangent vectors of $M$ i.e. $TM=\bigcup_{x\in M}T_xM$. For each $x\in U_\alpha$, the fibre $\pi^{-1}(x)$ of $x\in M$ is $T_xM\cong\{x\}\times\mathbb{R}^n$. The local trivialization map $h_\alpha:\pi^{-1}(U_\alpha)\longrightarrow U_\alpha\times\mathbb{R}^n$ is given by
$$h_\alpha(v)=(x,(v_\alpha^1,\cdots,v_\alpha^n)),\ v\in T_xU_\alpha(=T_xM),\ x\in U_\alpha.$$

A fibre bundle $(E,M,F,\pi)$ is called a vector bundle over $M$ if each fibre $F_x$ of $x\in M$ is a vector space. So, a tangent bundle is a vector bundle. The tangent bundle $TM$ is a differentiable manifold of dimension $2n$ with local coordinates in $\pi^{-1}(U_\alpha)$ being $(x_\alpha^1,\cdots,x_\alpha^n,v_\alpha^1,\cdots,v_\alpha^n)$. The Jacobian is given by
\begin{align*}
J(x_\beta^1,\cdots,x_\beta^n;x_\alpha^1,\cdots,x_\alpha^n)&=\frac{\partial(x_\beta^1,\cdots,x_\beta^n)}{\partial(x_\alpha^1,\cdots,x_\alpha^n)}\\
&=\begin{pmatrix}
\frac{\partial x_\beta^1}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^1}{\partial x_\alpha^n}\\
\vdots & \ddots & \vdots\\
\frac{\partial x_\beta^n}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^n}{\partial x_\alpha^n}
\end{pmatrix}:U_\alpha\cap U_\beta\longrightarrow\mathrm{GL}(n,\mathbb{R}).
\end{align*}
Let $g_{\alpha\beta}=J(x_\beta^1,\cdots,x_\beta^n;x_\alpha^1,\cdots,x_\alpha^n)$ Then $g_{\alpha\beta}$ satisfies
\begin{align*}
g_{\alpha\alpha}(x)&=I_n;\\
g_{\beta\alpha}(x)&=g_{\alpha\beta}^{-1}(x),\ x\in U_\alpha\cap U_\beta;\\
g_{\alpha\beta}(x)g_{\beta\gamma}(x)g_{\gamma\alpha}(x)&=I_n,\ x\in U_\alpha\cap U_\beta\cap U_\gamma.
\end{align*}
For $x\in U_\alpha\cap U_\beta$ and $f\in\mathbb{R}^n$, $h_{\alpha\beta}(x)(f)=g_{\alpha\beta}\cdot f$. So, $\mathrm{GL}(n,\mathbb{R})$ acts on the fibre $\mathbb{R}^n$. The map $g_{\alpha\beta}$ itself is often called a transition map.

If the transition map $h_\alpha\beta$ is the group action of a Lie group $G$ on the fibre $F$, the fibre bundle $(E,M,F,\pi)$ is called a $G$-bundle and the Lie group $G$ is called a structure group. The tangent bundle $TM$ is also a G-bundle with structure group $\mathrm{GL}(n,\mathbb{R})$.

In $\mathbb{R}^n$, there is a globally defined orthonormal frame
$$E_{1p}=(1,0,\cdots,0)_p,\ E_{2p}=(0,1,0,\cdots,0)_p,\cdots,E_{np}=(0,\cdots,0,1)_p.$$
For any tangent vector $X_p\in T_p(\mathbb{R}^n)$, $X_p=\sum_{i=1}^n\alpha^iE_{ip}$. Note that the coefficients $\alpha^i$ are the ones that distinguish tangent vectors in $T_p(\mathbb{R}^n)$. For a differentiable function $f$, the directional derivative $X_p^\ast f$ of $f$ with respect to $X_p$ is given by
$$X_p^\ast f=\sum_{i=1}^n\alpha^i\left(\frac{\partial f}{\partial x_i}\right).$$
We identify each $X_p$ with the differential operator
$$X_p^\ast=\sum_{i=1}^n\alpha^i\frac{\partial}{\partial x_i}:C^\infty(p)\longrightarrow\mathbb{R}.$$
Then the frame fields $E_{1p},E_{2p},\cdots,E_{np}$ are identified with
$$\left(\frac{\partial}{\partial x_1}\right)_p,\left(\frac{\partial}{\partial x_2}\right)_p,\cdots,\left(\frac{\partial}{\partial x_n}\right)_p$$
respectively. Unlike $\mathbb{R}^n$, we cannot always have a globally defined frame on a differentiable manifold. So it is necessary for us to use local coordinate neighborhoods that are homeomorphic to $\mathbb{R}^n$ and the associated frames $\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\cdots,\frac{\partial}{\partial x_n}$.

Example. The points $(x,y,z)$ are represented in terms of the spherical coordinates $(\phi,\theta)$ as
$$x=\sin\phi\cos\theta,y=\sin\phi\sin\theta,z=\cos\phi,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq 2\pi.$$
By chain rule, one finds the standard basis $\frac{\partial}{\partial\phi},\frac{\partial}{\partial\theta}$ for $T_\ast S^2$:
\begin{align*}
\frac{\partial}{\partial\phi}&=\cos\phi\cos\theta\frac{\partial}{\partial x}+\cos\phi\sin\theta\frac{\partial}{\partial y}-\sin\phi\frac{\partial}{\partial z},\\
\frac{\partial}{\partial\theta}&=-\sin\phi\sin\theta\frac{\partial}{\partial x}+\sin\phi\cos\theta\frac{\partial}{\partial y}.
\end{align*}
The frame field is not globally defined on $S^2$ since $\frac{\partial}{\partial\theta}$ at $\phi=0,\pi$. More generally, the following theorem holds.

Frame field on 2-sphere

Theorem. [Hairy Ball Theorem] If $n$ is even, a non-vanishing $C^\infty$ vector field on $S^n$ does not exist i.e. a $C^\infty$ vector field on $S^n$ must take zero value at some point of $S^n$.

The Hairy Ball Theorem tells us why we have ball spots on our heads. It can be also stated as “you cannot comb a hairy ball flat.” There may also be a meteorological implication of this theorem. It may implicate that there must be at least one spot on earth where there is no wind at all. No-wind spot may be the eye of a hurricane. So, as long as there is wind (and there always is) on earth, there must be a hurricane somewhere at all times.

It has been known that all odd-dimensional spheres have at least one non-vanishing $C^\infty$ vector field and that only spheres $S^1, S^3, S^7$ have a $C^\infty$ field of basis. For instance, there are three mutually perpendicular unit vector fields on $S^3\subset\mathbb{R}^4$ i.e. a frame field: Let $S^3=\{(x^1,x^2,x^3,x^4)\in\mathbb{R}^4: \sum_{i=1}^4(x^i)^2=1\}$. Then
\begin{align*}
X&=-x^2\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}+x^4\frac{\partial}{\partial x^3}-x^3\frac{\partial}{\partial x^4},\\
Y&=-x^3\frac{\partial}{\partial x^1}-x^4\frac{\partial}{\partial x^2}+x^1\frac{\partial}{\partial x^3}+x^2\frac{\partial}{\partial x^4},\\
Z&=-x^4\frac{\partial}{\partial x^1}+x^3\frac{\partial}{\partial x^2}-x^2\frac{\partial}{\partial x^3}+x^1\frac{\partial}{\partial x^4}
\end{align*}
form an orthonormal basis of $C^\infty$ vector fields on $S^3$.

A fibre bundle is an object $(E,M,F,\pi)$ consisting of

1. The total space $E$;
2. The base space $M$ with an open covering $\mathcal{U}=\{U_\alpha\}_{\alpha\in\mathcal{A}}$;
3. The fibre $F$ and the projection map $E\stackrel{\pi}{ \longrightarrow}M$.

The simplest case is $E=M\times F$. In this case, the bundle is called a trivial bundle. In general the total space may be too complicated for us to understand, so it would be nice if we can always find smaller parts that are simple enough for us to understand such as trivial bundles. For this reason, we want the fibre bundle to have the additional property: For each $U_\alpha\in\mathcal{U}$, there exists a homeomorphism $h_\alpha : \pi^{-1}(U_\alpha)\longrightarrow U_\alpha\times F$. Such a homeomorphism $h_\alpha$ is called a local trivialization. For each $x\in M$, $F_x:=\pi^{-1}(x)$ is homeomorphic to $\{x\}\times F$. $F_x$ is called the fibire of $x$.

Let $x\in U_\alpha\cap U_\beta$. Then $F_x^\alpha\subset\pi^{-1}(U_\alpha)$ and $F_x^\beta\subset\pi^{-1}(U_\beta)$ may not be the same. However, the two fibres are homeomorphic. For each $x\in M$, denote by $h_{\alpha\beta}(x)$ the homeomorphism from $F_x^\alpha$ to $F_x^\beta$. Then for each $x\in M$, $h_{\alpha\beta}(x)\in\mathrm{Aut}(F)$ where $\mathrm{Aut}(F)$ is the group of homeomorphisms from $F$ to itself i.e. the automorphism group of $F$. The map $h_{\alpha\beta}: U_\alpha\cap U_\beta\longrightarrow\mathrm{Aut}(F)$ is called a transition map. Note that for $U_\alpha,U_\beta\in\mathcal{U}$ with $U_\alpha\cap U_\beta\ne\emptyset$, $h_\alpha\circ h_\beta^{-1}:(U_\alpha\cap U_\beta)\times F\longrightarrow (U_\alpha\cap U_\beta)\times F$ satisfies
$$h_\alpha\circ h_\beta^{-1}(x,f)=(x,h_{\alpha\beta}(x)f)$$
for any $x\in U_\alpha\cap U_\beta$, $f\in F$

Tensors may be considered as a generalization of vectors and covectors. They are extremely important quantities for studying differential geometry and physics.

Let $M^n$ be an $n$-dimensional differentiable manifold. For each $x\in M^n$, let $E_x=T_xM^n$, i.e. the tangent space to $M^n$ at $x$. We denote the canonical basis of $E$ by $\partial=\left(\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^n}\right)$ and its dual basis by $\sigma=dx=(dx^1,\cdots,dx^n)$, where $x^1,\cdots,x^n$ are local coordinates. The canonical basis $\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^1}$ also simply denoted by $\partial_1,\cdots,\partial_n$.

Covariant Tensors

Definition. A covariant tensor of rank $r$ is a multilinear real-valued function
$$Q:E\times E\times\cdots\times E\longrightarrow\mathbb{R}$$
of $r$-tuples of vectors. A covariant tensor of rank $r$ is also called a tensor of type $(0,r)$ or shortly $(0,r)$-tensor. Note that the values of $Q$ must be independent of the basis in which the components of the vectors are expressed. A covariant vector (also called covector or a 1-form) is a covariant tensor of rank 1. An important of example of covariant tensor of rank 2 is the metric tensor $G$:
$$G(v,w)=\langle v,w\rangle=\sum_{i,j}g_{ij}v^iw^j.$$

In componenents, by multilinearity
\begin{align*}
Q(v_1\cdots,v_r)&=Q\left(\sum_{i_1}v_1^{i_1}\partial_{i_1},\cdots,\sum_{i_r}v_r^{i_r}\partial_{i_r}\right)\\
&=\sum_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}Q(\partial_{i_1},\cdots,\partial_{i_r}).
\end{align*}
Denote $Q(\partial_{i_1},\cdots,\partial_{i_r})$ by $Q_{i_1,\cdots,i_r}$. Then
$$Q(v_1\cdots,v_r)=\sum_{i_1,\cdots,i_r}Q_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}.\ \ \ \ \ \mbox{(1)}$$
Using the Einstein’s convention, (1) can be shortly written as
$$Q(v_1\cdots,v_r)=Q_{i_1,\cdots,i_r}v_1^{i_1}\cdots v_r^{i_r}.$$
The set of all covariant tensors of rank $r$ forms a vector space over $\mathbb{R}$. The number of components in such a tensor is $n^r$. The vector space of all covariant $r$-th rank tensors is denoted by
$$E^\ast\otimes E^\ast\otimes\cdots\otimes E^\ast=\otimes^r E^\ast.$$

If $\alpha,\beta\in E^\ast$, i.e. covectors, we can form the 2nd rank covariant tensor, the tensor product $\alpha\otimes\beta$ of $\alpha$ and $\beta$: Define $\alpha\otimes\beta: E\times E\longrightarrow\mathbb{R}$ by
$$\alpha\otimes\beta(v,w)=\alpha(v)\beta(w).$$
If we write $\alpha=a_idx^i$ and $\beta=b_jdx^j$, then
$$(\alpha\otimes\beta)_{ij}=\alpha\otimes\beta(\partial_i,\partial_j)=\alpha(\partial_i)\beta(\partial_j)=a_ib_j.$$

Contravariant Tensors

A contravariant vector, i.e. an element of $E$ can be considered as a linear functional $v: E^\ast\longrightarrow\mathbb{R}$ defined by
$$v(\alpha)=\alpha(v)=a_iv^i,\ \alpha=a_idx^i\in E^\ast.$$

Definition. A contravariant tensor of rank $s$ is a multilinear real-valued function $T$ on $s$-tuples of covectors
$$T:E^\ast\times E^\ast\times\cdots\times E^\ast\longrightarrow\mathbb{R}.$$ A contravariant tensor of rank $s$ is also called a tensor of type $(s,0)$ or shortly $(s,0)$-tensor.
For 1-forms $\alpha_1,\cdots,\alpha_s$
$$T(\alpha_1,\cdots,\alpha_s)=a_{1_{i_1}}\cdots a_{s_{i_s}}T^{i_1\cdots i_s}$$
where
$$T^{i_1\cdots i_s}:=T(dx^{i_1},\cdots,dx^{i_s}).$$
The space of all contravariant tensors of rank $s$ is denoted by
$$E\otimes E\otimes\cdots\otimes E:=\otimes^s E.$$
Contravariant vectors are contravariant tensors of rank 1. An example of a contravariant tensor of rank 2 is the inverse of the metric tensor $G^{-1}=(g^{ij})$:
$$G^{-1}(\alpha,\beta)=g^{ij}a_ib_j.$$

Given a pair $v,w$ of contravariant vectors, we can form the tensor product $v\otimes w$ in the same manner as we did for covariant vectors. It is the 2nd rank contravariant tensor with components $(v\otimes w)^{ij}=v^jw^j$. The metric tensor $G$ and its inverse $G^{-1}$ may be written as
$$G=g_{ij}dx^i\otimes dx^j\ \mbox{and}\ G^{-1}=g^{ij}\partial_i\otimes\partial_j.$$

Mixed Tensors

Definition. A mixed tensor, $r$ times covariant and $s$ times contravariant, is a real multilinear function $W$
$$W: E^\ast\times E^\ast\times\cdots\times E^\ast\times E\times E\times\cdots\times E\longrightarrow\mathbb{R}$$
on $s$-tuples of covectors and $r$-tuples of vectors. It is also called a tensor of type $(s,r)$ or simply $(s,r)$-tensor. By multilinearity
$$W(\alpha_1,\cdots,\alpha_s, v_1,\cdots, v_r)=a_{1_{i_1}}\cdots a_{s_{i_s}}W^{i_1\cdots i_s}{}_{j_1\cdots j_r}v_1^{j_1}\cdots v_r^{j_r}$$
where
$$W^{i_1\cdots i_s}{}_{j_1\cdots j_r}:=W(dx^{i_1},\cdots,dx^{i_s},\partial_{j_1},\cdots,\partial_{j_r}).$$

A 2nd rank mixed tensor may arise from a linear operator $A: E\longrightarrow E$. Define $W_A: E^\ast\times E\longrightarrow\mathbb{R}$ by $W_A(\alpha,v)=\alpha(Av)$. Let $A=(A^i{}_j)$ be the matrix associated with $A$, i.e. $A(\partial_j)=\partial_i A^i{}_j$. Let us calculate the component of $W_A$:
$$W_A^i{}_j=W_A(dx^i,\partial_j)=dx^i(A(\partial_j))=dx^i(\partial_kA^k{}_j)=\delta^i_kA^k{}_j=A^i{}_j.$$
So the matrix of the mixed tensor $W_A$ is just the matrix associated with $A$. Conversely, given a mixed tensotr $W$, once convariant and once contravariant, we can define a linear transformation $A$ such that $W(\alpha,v)=\alpha(A,v)$. We do not distinguish between a linear transformation $A$ and its associated mixed tensor $W_A$. In components, $W(\alpha,v)$ is written as
$$W(\alpha,v)=a_iA^i{}_jv^j=aAv.$$

The tensor product $w\otimes\beta$ of a vector and a covector is the mixed tensor defined by
$$(w\otimes\beta)(\alpha,v)=\alpha(w)\beta(v).$$ The associated transformation is can be written as
$$A=A^i{}_j\partial_i\otimes dx^j=\partial_i\otimes A^i{}_jdx^j.$$

For math undergraduates, different ways of writing indices (raising, lowering, and mixed) in tensor notations can be very confusing. Main reason is that in standard math courses such as linear algebra or elementary differential geometry (classical differential geometry of curves and surfaces in $\mathbb{E}^3$) the matrix of a linear transformation is usually written as $A_{ij}$. Physics undergraduates don’t usually get a chance to learn tensors in undergraduate physics courses. In order to study more advanced differential geometry or physics such as theory of special and general relativity, and field theory one must be able to distinguish three different ways of writing matrices $A_{ij}$, $A^{ij}$, and $A^i{}_j$. To summarize, $A_{ij}$ and $A^{ij}$ are bilinear forms on $E$ and $E^\ast$, respectively that are defined by
$$A_{ij}v^iv^j\ \mbox{and}\ A^{ij}a_ib_j\ (\mbox{respectively}).$$ $A^i{}_j$ is the matrix of a linear transformation $A: E\longrightarrow E$.

Let $(E,\langle\ ,\ \rangle)$ be an inner product space. Given a linear transformation $A: E\longrightarrow E$ (i.e. a mixed tensor), one can associate a bilinear covariant bilinear form $A’$ by
$$A'(v,w):=\langle v,Aw\rangle=v^ig_{ij}A^j{}_k w^k.$$ So we see that the matrix of $A’$ is
$$A’_{ik}=g_{ij}A^j{}_k.$$ The process can be said as “we lower the index $j$, making it a $k$, by mans of the metric tensor $g_{ij}$.” In tensor analysis one uses the same letter, i.e. instead of $A’$, one writes
$$A_{ik}:=g_{ij}A^j{}_k.$$ This is clearly a covariant tensor. In general, the components of the associated covariant tensor $A_{ik}$ differ from those of the mixed tensor $A^i{}_j$. But if the basis is orthonormal, i.e. $g_{ij}=\delta^i_j$ then they coincide. That is the reason why we simply write $A_{ij}$ without making any distiction in linear algebra or in elementary differential geometry.

Similarly, one may associate to the linear transformation $A$ a contravariant bilinear form
$$\bar A(\alpha,\beta)=a_iA^i{}_jg^{jk}b_k$$ whose matrix components can be written as
$$A^{ik}=A^i{}_jg^{jk}.$$

Note that the metric tensor $g_{ij}$ represents a linear map from $E$ to $E^\ast$, sending the vector with components $v^j$ into the covector with components $g_{ij}v^j$. In quantum mechanics, the covector $g_{ij}v^j$ is denoted by $\langle v|$ and called a bra vector, while the vector $v^j$ is denoted by $|v\rangle$ and called a ket vector. Usually the inner product on $E$
$$\langle\ ,\ \rangle:E\times E\longrightarrow\mathbb{R};\ \langle v,w\rangle=g_{ij}v^iw^j$$ is considered as a covariant tensor of rank 2. But in quantum mechanics $\langle v,w\rangle$ is not considered as a covariant tensor $g_{ij}$ of rank 2 acting on a pair of vectors $(v,w)$, rather it is regarded as the braket $\langle v|w\rangle$, a bra vector $\langle v|$ acting on a ket vector $|w\rangle$.