We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$ and a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}.$ In the following sections we make use of the Einstein summation convention introduced in [vector analysis](/en/physics/mathematical-physics/vector-analysis/curvilinear-coordinates/) and $\mathbb{K} = \mathbb{R} \lor\mathbb{K} = \mathbb{C}.$
## Definition
> *Definition 1*: a **tensor** is a multilinear mapping of the type
We refer to $\mathbf{T} \in \mathscr{T}_q^p(V)$ as a $(p, q)$-tensor; a mixed tensor of **contravariant rank** $p$ and **covariant rank** $q.$ It may be observed that we have $\dim \mathscr{T}_q^p (V) = n^{p+q}$ with $\dim V = n \in \mathbb{N}$.
It follows from definition 1 and by virtue of the isomorphism between $V^{**}$ and $V$ that $\mathbf{T} \in \mathscr{T}_1^0(V) = V^*$ is a covector and $\mathbf{T} \in \mathscr{T}_0^1(V) = V$ is a vector.
## Kronecker tensor
> *Definition 2*: let $\mathbf{k} \in \mathscr{T}_1^1(V)$ be the **Kronecker tensor** be defined such that
Let $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ and $\mathbf{v} = v^j \mathbf{e}_j \in V$ then the tensor properties and the definition of the Kronecker tensor imply that
> *Definition 3*: the outer product $f \otimes g: X \times Y \to \mathbb{K}$ of two scalar functions $f: X \to \mathbb{K}$ and $g: Y \to \mathbb{K}$ is defined as
>
> $$
> (f \otimes g)(x,y) = f(x) g(y),
> $$
>
> for all $(x,y) \in X \times Y$.
The outer product is associative, distributive with respect to addition and scalar multiplication, but not commutative.
Note that although the same symbol is used for the outer product and the denotion of a tensor space, these are not equivalent. But are closely related.
For the following statements we take $p=q=r=s=1$ without loss of generality.
> *Definition 4*: the mixed $(p, q)$-tensor $\mathbf{e}_i \otimes \mathbf{\hat e}^j \in \mathscr{T}_q^p(V)$ is defined as
For $\mathbf{T} \in \mathscr{T}^0_q(V)$ it follows that there exists holors $T_i \in \mathbb{K}$ such that $\mathbf{T} = T_i \mathbf{\hat e}^i$ with $T_i = \mathbf{T}(\mathbf{e}_i)$, are referred to as the **covariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
For $\mathbf{T} \in \mathscr{T}^p_0(V)$ it follows that there exists holors $T^i \in \mathbb{K}$ such that $\mathbf{T} = T^i \mathbf{e}_i$ with $T^i = \mathbf{T}(\mathbf{\hat e}^i)$, are referred to as the **contravariant components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
If $\mathbf{T} \in \mathscr{T}^p_q(V)$, it follows that there exists holors $T^i_j \in \mathbb{K}$ are coined the **mixed components** of $\mathbf{T}$ relative to a basis $\{\mathbf{e}_i\}$.
By definition tensors are basis independent. Holors are basis dependent.
> *Theorem 2*: let $\mathbf{S} \in \mathscr{T}^p_q(V)$ and $\mathbf{T} \in \mathscr{T}^r_s(V)$ be tensors with
> *Proposition 1*: the Gram matrix $G$ is symmetric and nonsingular such that
>
> $$
> g^{ik} g_{kj} = \delta^i_j,
> $$
>
> with $G^{-1} \overset{\text{def}}= (g^{ij})$.
??? note "*Proof*:"
Let $G$ be the Gram matrix, symmetry of $G$ follows from defintion 5. Suppose that $G$ is singular, then there exists $\mathbf{u} = u^i \mathbf{e}_i \in V \backslash \{\mathbf{0}\}$ such that $G \mathbf{u} = \mathbf{0} \implies u^i g_{ij} = 0$, as a result we find that
> with $\mathbf{g}(\mathbf{v}) = G \mathbf{v}$ for all $\mathbf{v} \in V$.
??? note "*Proof*:"
Let $\mathbf{u} \in V$ and let $\mathbf{\hat u} \in V^*$, suppose $\mathbf{\hat u}: \mathbf{v} \mapsto \bm{g}(\mathbf{u}, \mathbf{v})$ then we may define $\mathbf{g}: V \to V^*: \mathbf{u} \mapsto \mathbf{g}(\mathbf{u}) \overset{\text{def}} = \mathbf{\hat u}$.
Let $\mathbf{v} \in V \backslash \{\mathbf{0}\}: \mathbf{g}(\mathbf{v}) = \mathbf{0}$, then
for all $\mathbf{w} \in V$, which contradicts the non-degeneracy of the pseude inner product in definition 5. Hence $\mathbf{g}$ is injective, since $\dim V$ is finite $\mathbf{g}$ is also bijective.
Let $\mathbf{u} = u^i \mathbf{e}_i, \mathbf{v} = v^j \mathbf{e}_j \in V$ and define $\mathbf{g}(\mathbf{e}_i) = \text{g}_{ij} \mathbf{\hat e}^j$ such that
Since $u^i, v^j \in \mathbb{K}$ are arbitrary it follows that $\text{g}_{ij} = g_{ij}$.
Consequently the inverse $\mathbf{g}^{-1}: V^* \to V$ has the property $\mathbf{g}^{-1}(\mathbf{\hat u}) = G^{-1} \mathbf{\hat u}$ for all $\mathbf{\hat u} \in V^*$. The bijective linear map $\mathbf{g}$ is commonly known as the **metric** and $\mathbf{g}^{-1}$ as the **dual metric**.
It follows from theorem 3 that for $\mathbf{u} = u^i \mathbf{e}_i \in V$ and $\mathbf{\hat u} = u_i \mathbf{\hat e}^i \in V^*$ we have
Sofar, a vector space $V$ and its associated dual space $V^*$ have been introduced as a priori independent entities. An inner product provides us with an explicit mechanism to construct a bijective linear mapping associated with each vector by virtue of the metric.
