## 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 an alternating operator $\tilde f$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
\oplus_{i\in I} V\ar[r]^{\Lambda} \ar[rd]_f&\Lambda^I V\ar@{–>}[d]^{\tilde f}\\
& W
}
\end{xy}

**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., $$\tilde f(q(x))=\bar f(x)\forall x\in \otimes^I V.$$
\begin{xy}
\xymatrix{
\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
}
\end{xy}
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:
\begin{xy}
\xymatrix{
\oplus_{i\in I}V_i\ar[r]^{\otimes} \ar[dr]_f& \otimes_{i\in I} V_i\ar[d]^{\tilde f}\\
& W
}
\end{xy}
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:
\begin{xy}
\xymatrix{
V^k\ar[r]^{\Lambda} \ar[dr]_f & \Lambda^k(V)\ar[d]^{\tilde f}\\
& W
}
\end{xy}

**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 $\tilde f([z])=\bar f(z)$.
\begin{xy}
\xymatrix{
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
}
\end{xy}
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:
\begin{xy}
\xymatrix{
A\ar[r]^f & M(I)\\
I\ar[u]^i\ar[ru]_c & ,
}
\end{xy}
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