Ajunction from normed spaces to dual spaces

$\def \AA{A^{**}}\newcommand{\cC}{\mathbf{C\mbox{-}Vct}}

Lemma 1.
Suppose $\pi :V\to W$ is a continuous linear map between two normed spaces, then
the adjoint of $\pi $ defined by \begin{align}
\pi^*:&W^*\to V^*\notag\\
& f\mapsto f\circ \pi\label{adjoint}

is a continous linear map respect to norms of two spaces or waek* topology of two
spaces. Moreover, the following equation always holds: \begin{align}
\pi^{**}\circ i=i\circ \pi,

i.e., the following diagram is always commutative: \begin{equation}

Proof. Since \begin{equation}
\|f\circ \pi\|\leq\|f\|\|\pi\|,

Suppose $(f_i)$ is a net converges to $f$ in $W^*$ with respect to the weak* topology,
$$f_i(w)\to f(w) \forall w\in W.$$
For any $v\in V$, $\pi (v)\in W$, so
$$f_i(\pi (v))\to f(\pi (v)),$$
so $\pi ^*(f_i)$ weak*-converges to $\pi ^*(f)$.

For any $v\in V$, \begin{align*}
&(\pi^{**}\circ i)(v)(w’)=\pi^{**}(i(v))(w’)=(i(v)\circ \pi^*)(w’)\\
=&i(v)(\pi^*(w’))=i(v)(w’\circ \pi)=(w’\circ \pi)(v)\\
=&w’(\pi(v))=i(\pi(v))(w’)=(i\circ \pi)(v)(w’)~(w’\in W^*),

hence, $\pi ^{**}\circ i=i\circ \pi $. __

A normed space $M$ is said to be W-space if it admits a predual $^*M$, that is, $^*M$ is a normed
space and $M$ is isomorphic to $(^*M)^*$ as a normed space. Now Let $\cC $ and $\cW $ be two categories as

  • $\cC $: objects, a collection of normed spaces; arrows, all bouned linear maps
    between them.
  • $\cW $: objects, a collection of objects of form $V^*$ for some normed space $V$; arrows,
    all bouned linear maps between them that are also weak*-continuous.

Clearly, $\cW $ is a subcategory of $\cC $. Let $\ad $ assign to each bounded linear map $\pi :A\to B$ between two
normed space the double adjoint $\pi ^{**}: \AA \to \BB $, and $\id $ assign each bounded and weak*-continuous
map $\pi :M\to N$ between two W-spaces itself, then $\ad $ and $\id $ are two functors between $\cC $ and $\cW $.
Denotes the set of all arrows in $\cC $ from object $A$ to $B$ by $\cC (A,B)$, we have

Lemma 2.
for each object $A\in \cC $ and each obeject $M\in \cW $, there is a bijection of sets
&(\xymatrix{\ad(A)\ar[r]^\pi&M})\mapsto(\xymatrix{A\ar[r]^{\pi\circ i}&\id(M)}),

which is natural between two hom-functors

$$\hom _\cW (\ad (*),*), \hom _\cC (*,\id (*)): \cC ^{op}\times \cW \to \mathbf {Set}$$
as indicated by the following commutative diagram for each $\xymatrix {B\ar [r]^f & A}$ and $\xymatrix {M\ar [r]^g & N}$, \begin{align}
\cW(\ad A,M)\ar[d]_{((\ad f)^*,g_*)}\ar[r]^{n_{A,M}}&\cC(A,\id M)\ar[d]^{(f^*,(\id g)_*)}\\
\cW(\ad B,N)\ar[r]^{n_{B,N}}&\cC(B,\id N),

where $f^*$ is the operation turning an arrow $\pi :A\to B$ in $\cC $ into $f\circ \pi $, adn $g_*$ turning $\pi :M\to N$ in $\cW $ into $\pi \circ g$. In one
sentence, the triple $\langle \ad ,\id , n\rangle $ is an ajunction from $\cC $ to $\cW $.

By lemma 1, the following map is well-defined: \begin{align}
m=m_{A,M}:&\cC(A,\id M)\to\cW(\ad A,M)\notag\\
&(\xymatrix{A\ar[r]^{\pi}&M})\mapsto(\xymatrix{\AA\ar[r]^{(\pi^*\circ i)^*}&M})

For each $A\in \cC $ and each $M\in \cW $, we prove that $m_{A,M}$ is the inverse of $n_{A,M}$.

Pick an arrow $\xymatrix {\AA \ar [r]^{\pi ”} & M}$ in $\cW $. For any $a\in A$, \begin{align*}
&((\pi’’\circ i)^*\circ i)^*(i(a))(’m)=i(a)(((\pi’’\circ i)^*(i(’m)))=i(a)(i(’m)\circ (\pi’’\circ i))\\
=&(i(’m)\circ (\pi’’\circ i))(a)=i(’m)(\pi’’(i(a)))=\pi’’(i(a))(’m)~(’m\in^*M),


$$((\pi ”\circ i)^*\circ i)^*(i(a))=\pi ”(i(a)).$$
By Goldstine theorem (proposition V.4.1, [1]), $i(A)$ is weak* dense in $A^{**}$.
$$(\pi ”\circ i)^*\circ i)^* = \pi ”$$
since the maps of both sides are weak*-continuous.

Pick an arrow $\xymatrix {A\ar [r]^{\pi } & M}$ in $\cC $. For any $a\in A$, \begin{align*}
&((\pi^*\circ i)^*\circ i)(a)=(\pi^*\circ i)^*(i(a))=i(a)\circ (\pi^*\circ i)\\
=&(\pi^*\circ i)(a)=\pi^*(i(a))=i(a)\circ \pi=\pi(a).

$$(\pi ^*\circ i)^*\circ i=\pi .$$

Suppose $\pi \in \cW (\ad A, M)$. For each $\xymatrix {B\ar [r]^f & A}$ and $\xymatrix {M\ar [r]^g & N}$, \begin{align*}
\id(g)\circ n_{A,M}(\pi)\circ f=g\circ \pi\circ i\circ f,\\
g\circ n_{B,N}(\pi)\circ \ad(f)=g\circ \pi\circ f^{**}\circ i.

By \eqref{ad imbedding}, $\id (g)\circ n_{A,M}(\pi )\circ f=g\circ n_{B,N}(\pi )\circ \ad (f)$, hence the diagram \eqref{natural hom} is
commutative. __

By theorem IV.1.1, [2],
$$\eta _A:=n_{A,A}(\operatorname {1}_{\ad A})=i:A\to A^{**}$$
together with $\ad A$ is a universal arrow from $A\in \cC $ to the functor $\id :\cW \to \cC $ in the sense for any $M\in \cW $ and
any $\xymatrix { A\ar [r]^\pi & \id M}$ in $\cC $ there is an arrow $\xymatrix @C=1.75cm{\ad M\ar [r]^{\pi ”=n_{A,M}^{-1}(\pi )} & N}$ in $\cW $ filled in the commutative digram
$$\xymatrix { A\ar [r]^i\ar [rd]_\pi & \id \ad A\ar @{–>}[d]^{\id \pi ”} & \ad A\ar @{–>}[d]^{\pi ”=n_{A,M}^{-1}(\pi )} \\ & \id M. & M } $$
In other words,

Theorem 3. Each linear map $\pi :A\to M$ from a normed space $A$ and a W-space $M$ can be
extended to a unique linear map $(\pi ^*\circ i)^*:A^{**}\to M$ that are both bouded and weak*-continuous.


John B Conway. A course in functional analysis, volume 96. Springer
Science; Business Media, 2013.

Saunders MacLane. Categories for the Working Mathematician.
Springer-Verlag, New York, 1971. Graduate Texts in Mathematics, Vol. 5.

Exterior Power

**Definition**. Suppose $I$ is an index set, $V$ and $ W$ are linear vector spaces, a multilinear map $f:\oplus_{i\in I} V\to W$ is called an *alternating operator* if $f((v_i)_{i\in I})=0$ whenever $(v_i)_{i\in I}\in \oplus_{i\in I} V$ is a family of linearly dependent vectors in $V$, here $\oplus_{i\in I} V$ merely means a subspace of $\Pi_{i\in I} V$. The I-*exterior power* of a linear space $V$ with a linear space $\Lambda^I V$ with an alternating operator $\Lambda$ such that for any linear space $W$ and an alternating operator $f$, there exists a unique linear map $\tilde f$ such that the following diagram is commutative:
\oplus_{i\in I} V\ar[r]^{\Lambda} \ar[rd]_f&\Lambda^I V\ar@{–>}[d]^{\tilde f}\\
& W

**Existence**. Let $L$ be the linear subspace of $\otimes^I V$ gernerated by $\{v_1\otimes\cdots \otimes v_n |v_1,\cdots, v_n \mbox{ are linearly dependent. }\}$, then $\otimes^I V/L$ with $q\circ \otimes$ (where $q$ is the quotient map) is the I-*exterior power* of $V$.

*Proof*. For any alternating operator $f:V\to W$, we know that there exists some linear map $\bar f$ s.t. the left triangle commutative in the following diagram. Define $\tilde f$ as the map induced by $\bar f$, i.e., $$\tilde f(q(x))=\bar f(x)\forall x\in \otimes^I V.$$
\oplus_{i\in I} V\ar[r]^{\otimes} \ar[rd]_f&\otimes^I V\ar@{–>}[d]^{\bar f}\ar[r]^q&\Lambda^I V\ar@{–>}[dl]^{\tilde f}\\
& W
Since the image of $\tilde f$ is deterministic on each generator, $\tilde f$ is deterministic.

Infinite Tensor Product

**Definition**. Suppose $(V_i)_{i\in I}$ is a family of linear spaces. The tensor product of $(V_i)_{i\in I}$ is defined as a linear space $\otimes_{i\in I} V_i$ with a multilinear operator $\otimes:\oplus_{i\in I}V_i\to \otimes_{i\in I} V_i$ such that for any given linear space $W$ and a multilinear operator $f:\oplus_{i\in I}V_i\to W$, there is a unique linear operator $\tilde f:\otimes_{i\in I} V_i\to W$ such that the following diagram is commutative:
\oplus_{i\in I}V_i\ar[r]^{\otimes} \ar[dr]_f& \otimes_{i\in I} V_i\ar[d]^{\tilde f}\\
& W
What is $\oplus_{i\in I}V_i$? Just a subspace of the Cartesian product of $(V_i)_{i\in I}$.

**Existence**. Let $L(\oplus_{i\in I} V_i)$ denotes the set of the multilinear operators on $\oplus_{i\in I} V_i$. Define $\otimes_{i\in I} v_i\in L(L(\oplus_{i\in I} v_i))$ by $$\otimes_{i\in I} v_i(f)=f(\oplus_{i\in I} v_i),$$ and define $\otimes_{i\in I} V_i$ to be the subspace generated by $\{\otimes_{i\in I} v_i|v_i\in V_i\forall i\in I\}$ in $L(L(\oplus_{i\in I} V_i))$.
For any given multilinear operator $f:\oplus_{i\in I}V_i\to W$, in order to make the diagram commutative $\tilde f$ must be defined by $$\tilde f(\otimes_{i\in I}v_i)=f(\oplus_{i\in I}v_i).$$
And $\tilde f$ is also well-defined in this way:
Suppose $\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k= \sum_{k=1}^nl’_i\otimes_{i\in I}{v’}_i^k$.
For any $\tau\in L(W)$, $\tau \circ f\in L(\oplus_{i\in I}V_i)$, so
$\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k(\tau\circ f)=\sum_{k=1}^nl’_i\otimes_{i\in I}{v’}_i^k(\tau\circ f)$, i.e.,
$\tau(f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k))=\tau(f(\sum_{k=1}^nl’_i\oplus_{i\in I}{v’}_i^k))$, hence
$f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k)=f(\sum_{k=1}^nl’_i\oplus_{i\in I}{v’}_i^k)$.

**Tensor Algebra**. A character of an infinite set $S$ is that $\newcommand{\card}{\operatorname{card}}$ its cardinal number $\card(S)$ satisfies
$$\card(S)^2:=\card(S\times S)=\card(S).$$
So we can define a product in $\oplus_{C\leq \card(2^S)}\otimes_{i\in C} V.$
For $u_D\in \otimes_{i\in D} V, w_E\in \otimes_{i\in E} V$, define $$u_D\cdot w_E:=u_D\otimes w_E\in \otimes_{\card(D\sqcup E)} V,$$
and for $u=\oplus_{D\leq \card(2^S)} u_D$ and $w=\oplus_{E\leq \card(2^S)} w_E$, define $$u\cdot v:=\oplus_{C\leq \card(2^S)}\sum_{\card(D\sqcup E)=C} u_D\otimes w_E.$$
In order to let the sum in the above formula make sense, we can define $\oplus_i V_i=\{(v_i)|\{i|v_i\neq 0\} \mbox{ is a finite set. }\}$, or import norms.

Exterior Power

Given two vector spaces $V$ and $W$, an *alternating multilinear operator* from $V^k$ to $W$ is a multilinear map
$ f: V^k \to X $
such that whenever $v_1,\cdots,v_k$ are linearly dependent vectors in $V$, then
$f(v_1,\ldots, v_k)=0$.

The *kth exterior power* of a linear space $V$, is a linear space $\Lambda^k(V)$ with alternating multilinear operator $\Lambda:V^k\to \Lambda^k(V)$ such that for any given linear space and any alternating multilinear operator $f:V^k\to W$, there is a unique linear map $\tilde f$from $\Lambda^k(V)$ to $W$ such that $\tilde f\circ \Lambda=f$ as indicated by the following commutative diagram:
V^k\ar[r]^{\Lambda} \ar[dr]_f & \Lambda^k(V)\ar[d]^{\tilde f}\\
& W

**Construction**. Let $I$ be the linear subspace generated by $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$ in $\otimes^k V$. Then $\otimes^k V/I$ is a kth exterior power.

*Proof*. Let $\bar f:\otimes^k(V)\to W$ be the linear map such that $\bar f\circ \
\otimes= f$, $q:\otimes^k(V) \to \otimes^k V/I$ be the quotient map, and $\tilde f([z])=\bar f(z)$.
V^k\ar[r]^{\otimes} \ar[dr]_f & \otimes^k(V)\ar[d]^{\bar f}\ar[r]^q & \otimes^k(V)/I\ar[dl]^{\tilde f}\\
& W
If $z\in I$ then $z$ is a linear combination of element in $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$, so $\tilde f([z])=\bar f(z)= \sum_{i=1}^n k_if(x_1^i,\cdots,x_n^i) = 0$, so $\tilde f$ is well-defined.

>Then what about the infinite exterior power?

Universal Property of the Multiplier Algerba

We call an injective $*$-isomorphism $i:A\to B$ between two C*-algebra an *ideal homomorphism* if $i(A)$ is an closed ideal of $B$. The universal property of the multiplier algebra $M(I)$ of a C*-algebra $I$ is: for any C*-algebra $A$ and an ideal homomorphism $i:I\to A$ there is a unique $*$-homomorphism $f$ from $A$ to $M(I)$ such that the following diagram is commutative:
A\ar[r]^f & M(I)\\
I\ar[u]^i\ar[ru]_c & ,
where $c$ is the canonical ideal homomorphism.

*An application*. [If $I$ is a unital C*-algebra, then it cannot be an essential ideal of any other C*-algebra.][1]

[1]: https://math.stackexchange.com/q/3046342