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.

Tensor Algebra

**Definition**. Suppose $V$ is a vector space, the tensor algebra of $V$ is defined as an algebra $TV$ with a linear map $T:V\to TV$ such that for any algebra $A$ and any linear map $f:V\to A$ there exists a unique algebra homomorphism $\tilde f:TV\to A$ such that $f=\tilde f\circ T$ as indicated by the commutative diagram below.
\begin{xy}
\xymatrix{
V\ar[r]^T\ar[rd]_f & TV\ar@{–>}[d]^{\tilde f}\\
& A }
\end{xy}

**Existence**. $TV$ is defined as $\oplus_{n\in \mathbb N}\otimes^n V$ and $T$ is defined as a map send a vecctor $v$ to an assignment of $v$ to 1 and 0 to others.

*Proof*. Since $\tilde f\circ T(v)=f(v)$, $\tilde f(\otimes_{j}v_j)=\Pi_j f(v_j)$ where $\otimes_{j}v_j$ is an element of $\otimes^n V_j$, $f(\oplus_{i} w_i)=f(\sum_i w_i\chi_i)=\sum_i f(w_i)$. Hence $\tilde f$ is unique if it exists.

Let $\tilde f_i$ be the uqnique linear map as indicated by the following commutative diagram:

\begin{xy}
\xymatrix{
\Pi^i V\ar[r]^\otimes\ar[rd]_{\Pi^i f} & \otimes^i V\ar@{–>}[d]^{\tilde f_i}\\
&W}
\end{xy}
Define $\tilde f=\sum_i \tilde f_i .$ We only have to prove $\tilde f$ preserves product. $\tilde f((\oplus_n u_n)(\oplus_n v_n))=\tilde f(\oplus_n \sum_{m+l=n} u_m\otimes v_l)=\sum_n \sum_{m+l=n}\tilde f_n(u_m\otimes v_l)=\sum_n\sum_{m+l=n} \tilde f_m(u_m) \tilde f_l(u_l)=\sum_m \tilde f_m(u_m)\sum_l \tilde f_l(v_l)=\tilde f(\oplus_n u_n)\tilde f(\oplus_n v_n). $

How to Construct Bott Map

$\newcommand{\T}{\mathbb{T}}\newcommand{\R}{\mathbb{R}}$Suppose $p$ is a projection in $M_n(A^+)$. In order to get a unitary $u$ in $C(\T\to M_n(A^+))$, we only need a continuous function $f:\T\times \{0,1\} \to \mathbb {R}$ and to do the functional calculus on the second variable, that is, let $$u(z) =e^{if(z, p)}.$$
To make $u$ fall into $$K_1(SA)\cong \{[f]|f\in C(\T\to \mathcal U_n(A^+)), f(1)\sim_h 1\},$$
It is sufficient to set $$f(z,x)=|z-1|g(x),$$ where $g$ is a map from $\{0,1\}$ to $ \R$.

To make the map $B: p\mapsto u$ be invertible, we only need that the range of $e^{if(z,x)}$ does not equal to $\T$ and $g$ is invertible. Then the invert of $B$ is likely to be $B^{-1}: u\mapsto g^{-1}( \frac{1}{2 i}\operatorname{Ln} u(-1))$. Here we can simply take $g=\operatorname{id}$.

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?

the Relationship between Projections and Unitaries

Update. We can use functional calculus to obtain unitaries or projections.

—–

If $u$ is a unitary, then $\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & u^*\\ u &1 \end{array}\right)$ is a projection;

If $p$ is a projection, then $\left(\begin{array}{cc} p & 1-p\\ 1-p & p \end{array}\right)$ is a unitary.