How to Construct Bott Map


Infinite Tensor Product

**Definition**. Suppose $(V_i)_{i\in I}$ is a family of linear spaces. The direct sum of $(V_i)_{i\in I}$ is defined as $\oplus_{i\in I}V_i=\{(v_i)_{i\in I}|\{i|v_i\neq 0\}\mbox{ is a finite set }\}$, and 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} **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 $f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k)=f(\sum_{k=1}^nl'_i\oplus_{i\in I}{v'}_i^k)$. **Remark**. The proof seems the same as the finite case~

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.

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

What is “Free”

**Update**.Suppose $\mathbf{B}$ is a category, $\mathbf{A}$ is a subcategory of $\mathbf{B}$, and $i:X\to G$ is an injective morphism. If for each object $Y$ of $\mathbf{A}$ and each morphism $f:X\to Y$ there is a unique morphism $\tilde{f}:G\to Y$ such that the following diagram is commutative: \begin{xy} \xymatrix{ X\ar[r]^f \ar[d]_i&Y\\ G\ar[ur]_{\tilde{f}} & , } \end{xy} then $G$ is said to be a *free* object with a *basis* $X$. ------ Suppose $\mathbf{A}$ is a category, $G$ is an object of $\mathbf{A}$. If there is a parent category $\mathbf{B}$ of $\mathbf{A}$ with an object $X$ and an injective morphism $i:X\to G$ satisfying that for any object $Y$ of $\mathbf{A}$ and each morphism $f:X\to Y$ there is a unique morphism $\tilde{f}:G\to Y$ such that the following diagram is commutative: \begin{xy} \xymatrix{ X\ar[r]^f \ar[d]_i&Y\\ G\ar[ur]_{\tilde{f}} & , } \end{xy} then $G$ is said to be a *free* object. (discussed with S.Q.Huang)

Questions this week

- How to compute $V(\mathbb{B}/\mathbb{K})$. Suppose $\alpha:A\to B$ is surjective, is there a lift for each projection $p\in B$? - the limit of normal elements, the universal property of Borel functional calculus. - u^t

The isomorphism between $K_0$ groups under the unital case implies that under the non-unital case

This is a supplementary explanation for lemma 6.2.10, [[1](#wegge-olsen)]. The author states that to prove the map $a\mapsto \diag(a,0)$ from $A$ to $M_n(A)$ induces an isomorphism between $K_0(A)$ and $K_0(M_n(A))$, it suffices to prove that the map induces a isomorphism between $V(A)$ and $V(M_n(A))$. It is true for the unital case in which $K_0(A)$ is just the Grothendieck group of $V(A)$. But for the non-unital case? It can be proved by the unital case!(known from Dadarlat and Chung) Consider the following diagram: $\newcommand{\C}{\mathbb{C}}$ \begin{xy} \xymatrix{ 0\ar[r]\ar[d] & A\ar[r]^i\ar[d]^\alpha &A^+\ar@/^/[r]^{\pi}\ar[d]^{\alpha^+} &\C\ar[r]\ar[d]^{\alpha_{\C}}\ar@/^/ [l]^j &0\ar[d] \\ 0\ar[r] & M_n(A)\ar[r]^{i_n} &M_n(A^+)\ar@/^/[r]^{\pi_{ n}}&M_n(\C)\ar[r]\ar@/^/[l]^{j_n} &0 } \end{xy} (1)It is commutative. For example, $\alpha_\C\circ\pi(a,\lambda)=\alpha_\C(\lambda)=\diag(\lambda,0)$ and $\pi_n\circ\alpha^+(a,\lambda)=\pi_n((a,\lambda),0)=(\lambda,0)$ for each $(a,\lambda)\in A^+$, thus $\alpha_\C\circ\pi=\pi_n\circ\alpha^+$. (2)Its rows are split exact sequences of C*-algebras. It is easy to verify that $\operatorname{im}i_n=\ker \pi_n$ and $j_n\circ\pi_n=\operatorname{id}$. Since $K_0$ is a covariant functor that preserves split exactness (corollary 8.2.2, [[1](#wegge-olsen)]), the following diagram is also communicative and the rows are split exact too. \begin{xy} \xymatrix{ 0\ar[r]\ar[d] & K_0(A)\ar[r]^{i_*}\ar[d]^{\alpha_*} & K_0(A^+)\ar[r]^{\pi_*}\ar[d]^{\alpha^+_*} &K_0(\C)\ar[r]\ar[d]^{\alpha_{\C *}} &0\ar[d] \\ 0\ar[r] & K_0(M_n(A))\ar[r]^{i_{n*}} &K_0(M_n(A^+))\ar[r]^{\pi_{ n *}}&K_0(M_n(\C))\ar[r] &0 } \end{xy} From the unital case, we know that $\alpha^+_*:K_0(A^+)\to K_0(M_n(A^+))$ and $\alpha_{\C *}:K_0(\C)\to K_0(M_n(\C))$ are isomorphisms. Proof ends by applying the five-lemma to \begin{xy} \xymatrix{ 0\ar[r]\ar[d] & 0\ar[r]\ar[d] & K_0(A)\ar[r]^{i_*}\ar[d]^{\alpha_*} & K_0(A^+)\ar[r]^{\pi_*}\ar[d]^{\alpha^+_*} &K_0(\C)\ar[d]^{\alpha_{\C *}} \\ 0\ar[r] & 0\ar[r] & K_0(M_n(A))\ar[r]^{i_{n*}} &K_0(M_n(A^+))\ar[r]^{\pi_{ n *}}&K_0(M_n(\C)) } \end{xy} $1$ Wegge-Olsen, N. E. (1993). K-theory and C*-algebras.

a Positive Bilinear Map is Necessarily Bounded

Suppose $A$ and $B$ are Banach spaces, $C$ is a normed space, and $\sigma:A\times B\to C$ is a bilinear form. For each $a\in A$, denote the map $b\mapsto \sigma(a,b)$ by $\hat{a}:B\to C$; and for each $b\in B$ denote the map $a\mapsto \sigma(a,b)$ by $\hat{b}:A\to C$. **Lemma**. The bilinear form $\sigma$ is bounded is equivalent to $\hat{a}$ and $\hat{b}$ are both bounded for all $a\in A$ and $b\in B$. *Proof*. Since $\hat{b}$ is a bounded map on $A$ for all elment $b$ in the closed unit ball of $B$, $\sup_{a\in A_1}\|\hat{b}(a)\|\leq \|\hat{b}\|$, i.e., $\sup_{a\in A_1}\|\hat{a}(b)\|\leq \|\hat{b}\|$. By PUB, the family $\{\hat{a}|a\in A_1\}$ of bounded maps is bounded uniformly, that is, $\sup_{a\in A_1}\|\hat a\|\leq M$ for some $M>0$ . Hence, $\sup_{a\in A_1, b\in B_1}\|\sigma(a, b)\|=\sup_{a\in A_1, b\in B_1}\|\hat{a}(b)\|\leq M$. **Application**. Let $A$, $B$ and $C$ be C*-algebras.$\newcommand{\tensor}{\otimes}$ By a *positive* bilinear form $\sigma$ from $A\times B$ to $C$, we mean $\sigma(a,b)\geq 0$ for all positive elements $a$ in $A$ and $b$ in $B$, or equivalently, $\hat{a}$ and $\hat{b}$ are positive linear maps on $B$ and $A$ respectively. Since every positive linear map between two C*-algebra is bounded, by the above lemma, >a positive bilinear form in C*-algebras must be bounded. If $A$ and $B$ are C*-algebras, $\gamma$ is a C*-norm on $A\tensor B$, then the map $A\times B\to A\tensor_\gamma B$ is a positive bilinear form and thus is bounded, that is, $\gamma(a,b)\leq M\|a\|\|b\|$ for some positive number $M$.

Exercises on von Neumann Algebra


\begin{equation*} \lvert uv\rvert\leq \lVert u\rVert\lvert v\rvert. \end{equation*}

Indded,Let $v=w\lvert u\rvert$ and $uv=w'\lvert uv\rvert$ be the Polar Decomposition, then $\lvert uv\rvert^2=w'^*uw\lvert v\rvert$, thus \begin{align*} \lvert uv\rvert^2 & =\lvert v\rvert w^* u^* w' w'^*uw\lvert v\rvert\\ &\leq \lVert w'^*uw\rVert^2\lvert v\rvert^2\leq\l{u}^2\lvert v\rvert^2, \end{align*} Hence $\lvert uv\rvert\leq \lVert u\rVert\lvert v\rvert.$

$\textbf{Theorem}.$ Let $A$ be a $\vN$ on Hilbert space containing $\operatorname{id}_H$, then for any $\sigma$-continuous linear $\tau$ on $A$, there exists $u\in L^1(H)$ such that $$\tau(v)=\operatorname{tr}(uv)$$ for all $v\in A$.
