Lecture Notes

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====== 2020-01-29, Wednesday ======
$$\gdef\ot\otimes$$

===== Tensor power of a vector space =====
Let $V$ be a finite dimensional vector space. 

We denote the $k$-th tensor power of $V$ as 
$$ V^{\otimes k} = \underbrace{V\ot \cdots \ot V}_{\text{ $k$ times} } $$
Its elements are linear combinations of terms like $v_1 \otimes \cdots \ot v_k$, subject to the usual linearity relations. 

It is sometimes useful to consider the tensor algebra (we only mention it here, but do not use it later in this course).

** Definition (Tensor Algebra $T(V)$ ) **
$$T(V) = \R \oplus V \oplus V^{\ot 2} \oplus \cdots \oplus V^{\ot 3} \oplus \cdots $$
Given two elements $T = w_1 \ot \cdots \ot w_k$ and $T' = v_1 \ot \cdots \ot v_l$, their products is defined by juxtapostion.  
$$ T \ot T' = w_1 \ot \cdots \ot w_k \ot v_1 \ot \cdots \ot v_l $$


===== Exterior power of a vector space =====
** Definition (Exterior product $\wedge^k(V)$)**
The $k$-th exterior product $\wedge^k(V)$ is the vector space consisting of linear combinations of the following terms $v_1 \wedge \cdots \wedge v_k$, where the expression is linear in each slot, 
$$ c \cdot (v_1 \wedge \cdots \wedge v_k) = (c v_1) \wedge v_2 \wedge \cdots \wedge v_k $$
$$ (v_1+v_1') \wedge \cdots \wedge v_k = v_1 \wedge \cdots \wedge v_k + v_1'\wedge \cdots \wedge v_k$$
and the expression changes signs if we swap any two slots 
$$ v_1 \wedge \cdots \wedge v_i \wedge \cdots \wedge v_j\wedge \cdots \wedge v_k = -  v_1 \wedge \cdots \wedge v_j \wedge \cdots \wedge v_i\wedge  \cdots \wedge v_k, \forall 1 \leq i < j \leq k. $$

If $k=0$, we set $\wedge^0 V = \R$. If $k=1$, then $\wedge^1 V =V$. 

** Proposition ** If we choose a basis $e_1, \cdots, e_n$ of $V$, then for $0 \leq k \leq n$, the space $\wedge^k(V)$ has a basis consisting of the following vectors
$$ e_{i_1} \wedge \cdots \wedge e_{i_k}, \quad 1 \leq i_1 < i_2 < \cdots < i_k \leq n. $$

**Corrollary**
  * $\dim \wedge^k(V) = {n \choose k}$. 
  * If $k > n$, then $\wedge^k V = 0$. 

Just as we defined tensor algebra $T(V)$, we may define the exterior algebra $\wedge^* V$. This turns out to be very useful. 

** Definition(Exterior algebra $\wedge^* V$) ** 
$$ \wedge^* V := \bigoplus_{k=0}^{n} \wedge^k V, \quad \text{ where } \wedge^0 V:= \R.$$
The product between two elements is given by juxtaposition, more precisely, if $A = v_1 \wedge \cdots \wedge v_k \in \wedge^k V$, $B = w_1 \wedge \cdots \wedge w_l \in \wedge^l V$, then 
$$ A \wedge B :=  v_1 \wedge \cdots \wedge v_k \wedge w_1 \wedge \cdots \wedge w_l \in \wedge^{k+1} V.$$ 

===== Example of $\R^3$ =====
Let $V = \R^3$, and equip $V$ with the standard inner product. 

==== Cross product ====

Consider the $\wedge^2 V$, its dimension is ${3 \choose 2} = 3$, hence element of it are sometimes called pseudo-tensor. If we use the standard basis $e_1, e_2, e_3$ on $V$, then we have the following basis for $\wedge^2 V$: 
$$ e_1 \wedge e_2, \quad e_1 \wedge e_3, \quad e_2 \wedge e_3. $$


There is a bijection from $\wedge^2 V \to V$, called "Hodge star" $\star$, which goes as follows: 
$$ \star: e_1 \wedge e_2 \mapsto e_3, \quad e_2 \wedge e_3 \mapsto e_1, \quad e_3 \wedge e_1 \mapsto e_2. $$

Thus, we may recover our familiar cross-product $\b v \times \b w$ formula as following
$$ V \times V \xrightarrow{\wedge} \wedge^2 V \xrightarrow{\star} V. $$

//Exercise//: convince yourself that $\b v \wedge \b w = \star(\b v \wedge \b w)$. 

==== Volume Forms and Determinant ====
For $\wedge^3 V$, it is one-dimensional, with $e_1 \wedge e_2 \wedge e_3$ as a basis. Thus, element of $\wedge^* V$ are sometimes called pseudo-scalar. 

Given three vectors $v_1, v_2, v_3$, how to compute the signed volume formed by the parallelogram  $P(v_1, v_2, v_3)$ (skewed boxes) with sides $v_1, v_2, v_3$? 

From vector calculus, we know the answer is the determinant of the $3$ by $3$ matrix, whose column-vectors are $v_1, v_2, v_3$. 
$$ \text{ Volume of } P(v_1, v_2, v_3) = \det (v_1, v_2, v_3) = \det  \begin{pmatrix}
v_{11} & v_{21} & v_{31} \cr
v_{12} & v_{22} & v_{32} \cr
v_{13} & v_{23} & v_{33} 
\end{pmatrix}
$$

Now, we have another way to express it. 
$$ \text{ Volume of } P(v_1, v_2, v_3) = \frac{ v_1 \wedge v_2 \wedge v_3}{e_1 \wedge e_2 \wedge e_3} $$
Indeed, since both the numerator and denominators are elements of the one-dim vector space $\wedge^3 V$, the raio makes sense. 

===== Levi-Cevita Symbol and Kronecker Symbol =====
These are two symbols introduced in Boas's book. We list their definitions and some properties. 

The Kronecker symbol is used everywhere
$$ \delta_{ij} = \begin{cases} 
1 & \text{if } i=j \cr
0 & \text{if } i \neq j \cr
\end{cases} $$

The Levi-Cevita Symbol is a rank-3 tensor on $\R^3$. Its component with respect to the standard basis is
$$ \epsilon_{ijk} = \begin{cases} 
1 & \text{if } ijk=123,231,312 \cr
-1 & \text{if } ijk = 213,132,321 \cr
0 & \text{ if $ijk$ has repeated indices }
\end{cases} $$

For example, we can use $\epsilon_{ijk}$ to express the determinant
$$\det(v_1, v_2, v_3) = \sum_{ijk} \epsilon_{ijk} v_{1i} v_{2j} v_{3k}. $$

The Levi-Cevita symbol has generalization to higher dimension. It is a rank n tensor on $\R^n$, i.e, it has $n$ indices. Let $I$ denote the index. $\epsilon_I=1$ if $I$ can be obtained from $12 \cdots n$ by even number of permutation, and $\epsilon_I=-1$ if $I$ can be similarly obtained by odd number of permutations, and $\epsilon_I=0$ if there are repeated indices in $I$. 

Another useful property is that
$$ \sum_i \epsilon_{ijk} \epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm} \delta_{kl} $$
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