Calculus III - Lecture 3

I cannot claim to have a summary of the whole lecture, that’s why I switched the wording to ‘highlights’

Lecture highlights:

1. tensor algebra
2. graded vector space
3. determinants

M: Do abstract nonsense, then start explaining that algebra has bilinear operation and what is a bilinear operation for an entire minute
B: Beuh

He literally defined notations in the second lecture and proceed to ignore it.

M: This will probably not be used in this course.

M: These stuff are probably too difficult for this course.

Not much non-mathematical content this time. To make up for it, below is a half complete summary of the lecture. I am just Chinese roomming (search Chinese room) and sometimes I don’t even understand my own notes so I draw my notes using a commutative diagram software. Note that sometimes it is not a commutative diagram and just a rough pictorial transcription of my notes.

All vector spaces below are finite dimensional.

A tensor product is a vector space $V_1\otimes \cdots \otimes V_k$ universal multilinear thing $\iota$ as defined below:

where $Z$ is any vector space.

There is a proof that all such universal maps $u$ are naturally identifiable by general nonsense, where the maps below are all bilinear:

A tensor algebra is a vector space equipped with a multilinear product $\otimes$.

We have

$$T^{0}V:=\mathbb{F}$$ $$T^{1}V:=V$$ $$T^{k}V:=\underset{i=1}{\overset{k}{⨂}}{V}_i$$

We have a $\mathbb{Z}$-graded vector space

$$T^{\bullet }V:=\underset{k\in \mathbb{Z}}{⨁}{T}^{k}V.$$

Where does the negative grading come from? I have no idea.

The $k$-th homogeneous piece is $T^{k}V$.

Example (the polynomial space):

$$\mathbb{R}[t]=\underset{k=0}{\overset{\mathrm{\infty }}{⨁}}\mathbb{R}{t}^k$$

We claim that $T^{\bullet }$ is associative. The proof is as follows: (I have no idea what happened my notes are too messy)

The ideal of a vector space is a subspace, i.e. closed under linear combination. The ideal of an algebra is additionally closed under the bilinear product.

Let’s consider the symmetric tensor algebra $S^{\bullet }V$. This is also a $\mathbb{Z}$-graded algebra.

Modding by this ideal forces $u\cdot v=v\cdot u$, where $\cdot$ denotes our symmetric tensor product.

Now we consider the anti-symmetric tensor, or exterior, algebra ${\mathrm{\Lambda }}^{\bullet }V$.

There is also the Heisenberg and Clifford algebra, which are quantizations.

The cross product satisfies the Jacobean identity

Now we consider Lie algebras. ${\mathcal{U}}_{\mathfrak{g}}$ is the universal enveloping algebra of the Lie algebra $\mathfrak{g}$, defined as a quotient.

This breaks the $Z$ grading due to the ideal consisting of 2 grades.

Similarly, many algebras we consider are quotients of the tensor algebras.

The basis of $U\otimes V$ is the minimal spanning set of $u_i\otimes v_j$, where $u_i\in {\mathcal{B}}_U$ and $v_j\in {\mathcal{B}}_V$. It has dimension $mn$, where $m=\dim U$ and $n=\dim V$.

In the algebra $S^{\bullet }V$, the minimal spanning set of $S^{k}V$ is $v_{i_1}\cdot v_{i_2}\cdot \cdots \cdot v_{i_k}:1\le i_1\le \cdots \le i_k\le n$, which has a dimension that Prof. Meng was too lazy to calculate, and I am also too lazy to calculate it.

We have a similar case for ${\mathrm{\Lambda }}^{k}V$, this has dimension $(\genfrac{}{}{0ex}{}{n}{k})$. These dimensions can be calculated by generating functions:

$$\sum _{k=0}^n\dim {\mathrm{\Lambda }}^{k}V=(1+t{)}^{\dim V},$$ $$\sum _{k=0}^{\mathrm{\infty }}\dim S^{k}V=(1-t{)}^{-\dim V},$$

This negative exponent means you have to expand to obtain its power series.

Actually it is interesting how it is $\mathrm{\infty }$.

We define the determinant of a vector space $V$ to be

$$detV:={\mathrm{\Lambda }}^{\dim V}.$$

This is also called the determinant line of $V$ because it is one dimensional.

The below wedge product is nonzero because $b_i$ form basis and so are linearly indepedent by definition.

An orientation of a vector space is an equivalence class of ${\mathcal{B}}_V$. By fixing an origin on its determinant line, and mapping the wedge product of a chosen basis to an element on either side of it.

$(v_1,\dots ,v_n)\sim (u_1,\dots ,u_n)$ if and only if $v_1\wedge \cdots \wedge v_n=c\mathrm{\Lambda }{u}_i$ for some positive $c$.

$$\underset{i=1}{\overset{n}{⨂}}{v}_i\equiv {\mathrm{Map}}^{\mathsf{ML}}({\times }_{i=1}^n{V}_i,\mathbb{F})$$

We also have the below natural equivalences:

$$\mathrm{Hom}(U,\mathrm{Hom}(V,W))\equiv \mathrm{Hom}(U\otimes V,W),$$ $$\mathrm{End}(V)\equiv V^{\ast }\otimes V.$$

Get ready for the epic moment of this lecture.

So the trace of an endomorphism is actually equivalent to $I$.

This concludes the discussion about tensor algebras. There is a little treat for sticking with us.

Consider an $n\times n$ matrix $\mathbf{A}$. We have an expansion for a very important polynomial

$$det(\mathbf{I}+t\mathbf{A})=1+(\mathrm{tr}\mathbf{A})t+(\frac{(\mathrm{tr}\mathbf{A}{)}^2}{2!}-\frac{\mathrm{tr}({\mathbf{A}}^2)}{2})t^2+(\frac{(\mathrm{tr}\mathbf{A}{)}^3}{3!}-\frac{(\mathrm{tr}{\mathbf{A}}^{\mathbf{2}})\mathrm{tr}\mathbf{A}}{2}+\frac{\mathrm{tr}({\mathbf{A}}^3)}{3})+\cdots$$

This is computed with Feynman diagrams.

I actually don’t know why the constant term is $1$ and not $det\mathbf{A}$. Maybe he assumed the determinant is one.

We will talk about affine spaces next lecture.