Underlying Structure and Necessary Assumption in Linear Transformations

```math
\def\sub#1#2{{#1}_{#2}}
\def\ds{\displaystyle}
\def\bm#1{\boldsymbol{#1}}
\def\hom#1{\mathrm{Hom}(#1)}
\newcommand{\ket}[1]{|#1\rangle}
\newcommand{\bra}[1]{\langle#1|}
\newcommand{\bk}[2]{\langle#1|#2\rangle}
```
In this post we explain an underlying structure we always overlook when we use matrices. After that, we compare two ways of understanding toward the transformation law of linear transformations (by applying the structure), which is useful in quantum mechanics and other linear phenomena.

# Introduction
In order to deal with problems intuitively, when we use matrices, especially the finite dimensional ones, we usually show the matrices as a group of numbers configure a rectangle. For instance, we are familiar with the representation of rotation matrix on $\mathbf{R}^2$:$$R_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\\\sin\theta&\cos\theta\end{pmatrix}.$$When using this notation, we are actually implicitly allowing the representation of matrices by **tensor products**:$$A=\sum_{i,j}\sub{a}{ji}\\:\sub{\bm{e}}{i}^\mathrm{T}\otimes\sub{\bm{e}}{j}$$Here, $A$ is a linear map from vector space $V$ to $W$, and $\sub{a}{ji}\in\mathbf{K},\\;\sub{\bm{e}}{i}^\mathrm{T}\in V^\ast,\\;\sub{\bm{e}}{j}\in W$. That is to say, each slot in the matrix represents a tensor product composed by one of the bases in both $V^\ast$ (dual space of $V$) and $W$, or in other words, the matrix is expanded by these two bases. Under this notation, it is natural for us to define the product between matrices (or furthermore, tensors) by defining the bi-orthogonality property $$\sub{\bm{e}}{k}^\mathrm{T}\sub{\bm{e}}{l}=\sub{\delta}{kl}.$$We should pay attention to that this notation is not trivial: when we refer to a linear transformation, we don't choose particular bases in the vector spaces. The true background of this, is a **canonical isomorphism** between vector space $V^\ast\otimes W$ and $\hom{V,W}$.

# Review: Dual Space and Inner Product
Before further explanation of the canonical isomorphism above, let's review the definition of dual space and inner product. Readers who are familiar with those definition can skip this part.

Consider a vector space $V$ on field $\mathbf{F}$. The **(algebraic) dual space** (denoted by $V^\ast$) is defined as the set of all linear maps $\psi: V\to \mathbf{F}$. In this sense of meaning, $\psi$ is a linear functional satisfying $$\forall x\in V,\\;a\in\mathbf{F}\quad\begin{cases}(\psi_1+\psi_2)(x)=\psi_1(x)+\psi_2(x)\\\\(a\psi)(x)=a(\psi(x))\end{cases}$$Remember that we have not defined the inner product yet; the value of $\psi(x)$ can be seen as a type of product with no appealing properties. Inner product, such as the dot product for Euclidean space, is defined in the manner below:$$\langle\cdot,\cdot\rangle:V\times V^\ast\to\mathbf{K}$$satisfies
1. Conjugate symmetry: $\langle x,y\rangle=\overline{\langle y,x\rangle}.$
2. Linearity in the first argument: $\forall a,b\in\mathbf{K}\\;\langle ax+by,z\rangle=a\langle x,z\rangle+b\langle y,z\rangle.$
3. Positive-definiteness: $\langle x,x\rangle\ge0.$

Here the most important difference between these two map is that inner product can be regarded as an inner property of vector space $V$. When we think of the structure of $V$-related spaces, we can take inner product in account.

# Canonical Isomorphism
## The Detailed Form of Map 
Now we return to the canonical isomorphism[^1]. Consider the map below:$$\begin{aligned}&\phi:V^\ast\otimes W\to\hom{V,W},\\\\&\phi(v'\otimes w):V\to W,\\\\&\phi(v'\otimes w)(v)=v'(v)w.\end{aligned}$$Here we denote $v'\in V^\ast$ as a linear functional maps $V$ to $\mathbf{F}$. Since all the linear transformation from $V$ to $W$ construct $\hom{V,W}$, the isomorphic and bi-linear property is apparent.

The whole setting of this map, just as what we have claimed in the introduction, is along with the hope that we can express the matrix in a more intuitive way. The multiplication and the expansive notation of matrix **obey the rule** (the map) shown above.

# Why can't it be V?
We have shown that $V^\ast\otimes W\cong \hom{V,W}$. However, another question comes up: Can't it be simpler? The dual space $V^\ast$ is still not intuitive enough. Why can't we just replace it with $V$? The necessity of $V^\ast$ turns out to be in canonicality.

It is true that there is a isomorphism $V\otimes W\cong V^\ast\otimes W$, but not canonical in general. Indeed, to claim that $V\cong V^\ast$, you need to *mention a specific class of basis* in $V$ in order to invoke basis in $V^\ast$. The only exception is that $V$ is an inner product space; with the existence of inner product, we can say that structually $V\cong V^\ast$ is a canonical isomorphism. Therefore, for example in the Euclidean space with dot product, $V\otimes W\cong \hom{V,W}$ is a canonical isomorphism, but we can't easily simplify the canonical isomorphism for any arbitrary vector spaces.

# Application: Transformation for Matrices
In quantum mechanics, we often confront some linear transformations for operators, such as the Heisenberg picture and parity transformations. In these examples we see a common method, that is, to transform an operator we need to *interpose it in the middle*:$$\hat{\mathcal{O}}\mapsto e^{i\hat{\mathscr{H}}t/\hbar}\\:\hat{\mathcal{O}}\\:e^{-i\hat{\mathscr{H}}t/\hbar},\qquad\hat{\mathcal{O}}\mapsto\hat{\pi}\\:\hat{\mathcal{O}}\\:\hat{\pi}^\dagger$$Although we can understand this weird form by considering the replacement of bases in tensor products, we can also justify this form by *maintaining the isomorphic structure* explained above (despite the fact that these two are intrinsically equivalent).

:::note warn
Notation Change
In this section, we change the order of tensor products: $$\hat{\mathcal{O}}=\sum_{i,j}\sub{o}{ji}\ket{j}\otimes\bra{i}$$where $\sub{o}{ji}\in\mathbf{C},\\;\ket{j}\in\mathcal{H},\bra{i}\in\mathcal{H}^\ast$. This change is due to the incorrespondence between the math-like notations and physics-like notations (same for inner product $\bk{i}{j}$).
:::
## Replacement of Bases
Using the tensor product expansion (or decomposition) (equivalent to apply the isomorphic map $\phi$), we can denote operator $\hat{\mathcal{O}}$ as $$\hat{\mathcal{O}}=\sum_{i,j}\sub{o}{ji}\ket{j}\otimes\bra{i}.$$From this, we can calculate the operator transformed by operator $\hat{g}$ as $$\hat{g}\\:\hat{\mathcal{O}}\\:\hat{g}^\dagger=\hat{g}\left(\sum_{i,j}\sub{o}{ji}\ket{j}\otimes\bra{i}\right)\hat{g}^\dagger=\sum_{i,j}\sub{o}{ji}(\hat{g}\ket{j})\otimes(\bra{i}\hat{g}^\dagger)$$which clearly shows that the transformation is equivalent to **change in bases** $\ket{i}\mapsto\hat{g}\ket{i}$. This corresponds with our cognition toward transformations in vectors (bras) invoked by operator $\hat{g}$, so it is easy to understand.

## Another Perspective From Group Action
If we regard the set of all linear operators (no matter they have actual physical meaning or not) as a **linear transformation group** $G:=\hom{\mathcal{H},\mathcal{H}}$, we can consider a **group action** from $\mathcal{H}$ (resp. $\mathcal{H}^\ast$) to $\mathcal{H}$ (resp. $\mathcal{H}^\ast$) itself: $$\begin{aligned}&\Phi:G\times \mathcal{H}\to \mathcal{H},\quad\Phi(g,\ket{x})=\ket{g\cdot x}=\hat{g} \ket{x}\\\\&\Phi:G\times \mathcal{H}^\ast\to \mathcal{H}^\ast,\quad\Phi(g,\bra{x})=\bra{g\cdot x}=\bra{x}\hat{g}^\dagger\end{aligned}$$

Here, about the action, we define some structures naturally derived from quantum mechanics:
1. Duality: $\bk{x'}{g\cdot x}=\bra{x'}\hat{g}\ket{x}=\left[\hat{g}^\dagger\ket{x'}\right]^\dagger\ket{x}=\bk{g^\dagger\cdot x'}{x}.$
1. tensor product: $\hat{g}(\ket{x}\otimes\ket{y})=\hat{g}\ket{x}\otimes\hat{g}\ket{y}=\ket{g\cdot x}\otimes\ket{g\cdot y}$
1. Action toward operator: $\forall \mathcal{O}\in G,\\;\widehat{g\cdot \mathcal{O}}\ket{x}=\hat{g}\\:\hat{\mathcal{O}}\\:\hat{g}^\dagger\ket{x}=\hat{g}\\:\hat{\mathcal{O}}\ket{g^\dagger\cdot x}$

The last one is the exact statement we want to justify.

In principle, we expect operator $\hat{g}$ to act *in the same way* toward two isomorphic Hilbert spaces. So naturally the statement below holds true: $$g\cdot\phi(u)=\phi(g\cdot u)$$ Every isomorphism satisfies the equation above is namely called **$G$-isomorphism**.

Let's check if the relationship $$\widehat{g\cdot\phi(\ket{y}\otimes\bra{x'})}\ket{x}=\widehat{\phi(g\cdot(\ket{y}\otimes\bra{x'}))}\ket{x}$$ truly stands or not. Using the three definitions above, the left-hand side can be calculated as $$\begin{aligned}\widehat{g\cdot\phi(\ket{y}\otimes\bra{x'})}\ket{x}&=\hat{g}\\:\widehat{\phi(\ket{y}\otimes\bra{x'})}\ket{g^\dagger\cdot x}\\\\&=\hat{g}\\:\bk{x'}{g^\dagger\cdot x}\ket{y}\\\\&=\bk{g\cdot x'}{x}\ket{g\cdot y},\end{aligned}$$and samely right-hand side can be calculated as $$\begin{aligned}\widehat{\phi(g\cdot(\ket{y}\otimes\bra{x'}))}\ket{x}&=\widehat{\phi(\ket{g\cdot y}\otimes\bra{g\cdot x'})}\ket{x}\\\\&=\bk{g\cdot x'}{x}\ket{g\cdot y},\end{aligned}$$ so indeed they correspond.

From this calculation, we can learn that the form of transformation for operators is necessary for *keeping the isomorphic structure* $\mathcal{H}^\ast\otimes\mathcal{H}\cong \hom{\mathcal{H},\mathcal{H}}$ while considering the action of any operator $g\in \hom{\mathcal{H},\mathcal{H}}$, which allows the expansive notation in the former way of understanding. This interpretation is better in the sense of *circumventing the difficulty* dealing with infinite dimensional descriptions.

[^1]: Canonical isomorphism is often referred as the only isomorphism between two groups in group theory. In vector spaces, canonical isomorphism is an isomorphism that doesn't depend on the bases in the vector spaces.