Looking for opportunities for further study

I’m looking for opportunities for further study(PhD program). The following is my CV. If you are interested, do not hesitate to contact me( email: cding@mail.nankai.edu.cn ).

$z(M)M$ is atomic

Suppose $M$ is a W*-algebra, $z$ is the supremum of all minimal projections in $M$. It is well-known that $z$ is a central porjection in $M$. Let $M_0=zM$, then $M_0$ is also a W*-algebra. Denote the collection of all minimal projections in a W*-algebra $M$ by $\mathscr P_m^M$.

**Lemma**. $\mathscr P_m^M=\mathscr P_m^{M_0}$.

*Proof*. Since $m=zm$ for any minimal projection $m$ in $M$, $\mathscr P_m^M\subset\mathscr P_m^{M_0}$. Conversely, suppose $m_0$ is a minimal projection in $M_0$, $p\in M$ and $p\leq m_0$, then $zp\leq zm=m$, hence $zp=0$ or $zp=m_0$. Therefore, $m_0p=m_0zp$ equals to $0$ or $m_0$, i.e., $m_0$ is a minimal projection in $M$.

Now, we can denote $\mathscr P_m^M$ and $\mathscr P_m^{M_0}$ by the same notation $\mathscr P_m$.

**Theorem**. $M_0$ is atomic, that is, each projection in $M_0$ dominates a minimal projection.

*Proof*. Suppose $p$ is a non-zero projection in $M_0$, then $p\leq z$, so, there is a minimal projection $m$ such that $pm\neq 0$( otherwise, $z=\sup\{m\in \mathscr P_m\}\leq 1-p$, $pz=0$ ), which is equivalent to $pmp\neq 0$. Since $$pmpMpmp\subset pmMmp=p(\mathbb{C}m) p=\mathbb{C}pmp,$$ $(pmp)^2$ is a scalar multiple of $pmp$. Let $m_p=pmp/\|pmp\|$, then $m_p$ is a projection and $m_pMm_p= \mathbb C m_p$, that is, $m_p$ is a minimal projection. What’s more, $m_p\leq p$ for $m_pp=pm_p=m_p$.

**Theorem**. For any $q$ in $M_0$, we have $q=\sup\{m\in \mathscr P_m|m\leq q\}$.

*Proof*. Denote $\sup\{m\in \mathscr P_m|m\leq q\}$ by $q_0$, then $q_0\leq q$ and $q_0\in M_0$. If $q_0\neq q$, then $q-q_0$ dominates a minimal projection, say $m$. So, $m\leq q$, and thus $m\leq q_0$ by the definition of $q_0$. Therefore, $m(q-q_0)=mq-mq_0=m-m=0$, a contradiction to $m\leq q-q_0$.

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 $p\in B$?

– Is there any topology s.t. $f_n\to f$ is equivalent to $f_n$ converges uniformly to $f$ on any compact set?

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.

$\int_X F\operatorname{d} \mu=\int_Y F\circ g\operatorname{d} \nu $ where $g$ is invertible and $ \nu(B)=\mu(g B)$

Suppose $g$ is a measurable map from a measure space $(X,\mathscr F,\mu)$ to a measurable space $(Y,\mathscr S)$. Define $$\nu(B)=\mu(g^{-1} B) \forall B\in \mathscr S,$$ then $(Y,\mathscr S,\nu)$ is a measure space, and for any measurable function $f$ on $Y$, the following equation holds as long as one side makes sense:$\newcommand{\d}{\operatorname{d}}$
$$\int_Y f\d \nu = \int_X f\circ g\d \mu. $$

In particular, if $g$ is invertible and $h=g^{-1}$, then $$\nu(B)=\mu(h B)\forall B\in \mathscr S,$$ and for any measurable function $F$ on $X$, the following equation holds as long as one side makes sense:$\newcommand{\d}{\operatorname{d}}$
$$\int_X F\d \mu=\int_Y F\circ h\d \nu . $$