an update of various things

docs/mathematics/linear-algebra/tensors/tensor-formalism.md

@@ -45,7 +45,7 @@ $$
 
 ## Outer product
 
-> *Definition 3*: the outer product $f \otimes g: X \times Y \to F$ of two scalar functions $f: X \to F$ and $g: Y \to F$ is defined as
+> *Definition 3*: the **outer product** $f \otimes g: X \times Y \to F$ of two scalar functions $f: X \to F$ and $g: Y \to F$ is defined as
 >
 > $$
 >   (f \otimes g)(x,y) = f(x) g(y),
@@ -133,9 +133,45 @@ By definition tensors are basis independent. Holors are basis dependent.
 
 We have from theorem 2 that the outer product of two tensors yields another tensor, with ranks adding up.
 
+## Interior product
+
+> *Definition 5*: the **left interior product** $f \llcorner \alpha: Y \to F$ of a scalar function $f:X \times Y \to F$ and a scalar $\alpha \in X$ is defined as
+>
+>   $$
+>       (f \llcorner \alpha)(y) = f(\alpha,y) \qquad \forall y \in Y,
+>   $$
+>
+> and the **right interior product** $f \lrcorner \alpha: X \to F$ for a scalar $\alpha \in Y$ is defined as
+>
+> $$
+>       (f \lrcorner \alpha)(x) = f(x,\alpha) \qquad \forall x \in X.
+> $$
+
+Note that neither interior product is associative, commutatitive and only distributive in the field addition. 
+
+With the interior product we can partially contract a tensor (i.e. reduce its basis) with a (co)vector. Consider $\mathbf{T} \in \mathscr{T}^2_0(V)$ then
+
+$$
+    \begin{align*}
+        \mathbf{T} \llcorner \mathbf{\hat v} &= T^{ij} \mathbf{k}(\mathbf{\hat v}, \mathbf{e}_i) \mathbf{e}_j,\\
+        &= v_i T^{ij} \mathbf{e}_j.
+    \end{align*}
+$$
+
+or similarly
+
+$$
+    \begin{align*}
+        \mathbf{T} \lrcorner \mathbf{\hat v} &= T^{ij} \mathbf{e}_i \mathbf{k}(\mathbf{\hat v}, \mathbf{e}_j),\\
+        &= v_j T^{ij} \mathbf{e}_i.
+    \end{align*}
+$$
+
+for all $\mathbf{\hat v} \in V^*$. 
+
 ## Inner product
 
-> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to F$ which satisfies
+> *Definition 6*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to F$ which satisfies
 >
 > 1. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u}),$ 
 > 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in F: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
@@ -144,7 +180,7 @@ We have from theorem 2 that the outer product of two tensors yields another tens
 
 It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existence of an inner product is never implied. 
 
-> *Definition 6*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as 
+> *Definition 7*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as 
 > 
 > $$
 >   g_{ij} = \bm{g}(\mathbf{e}_i, \mathbf{e}_j).
@@ -227,7 +263,7 @@ $$
 
 with $u^j = g^{ij} u_i$. 
 
-> *Definition 7*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by 
+> *Definition 8*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by 
 > 
 > $$
 > \mathbf{g}^{-1}(\mathbf{\hat e}^i) = g^{ij} \mathbf{e}_j.