[proofplan] We form the [tensor product](/page/Tensor%20Product) distribution $u \otimes v$ on $X \times X$ and use the tensor product wave front set estimate to locate its singular directions. The product $uv$ is then the pullback of this tensor product along the diagonal map $\Delta: X \to X \times X$, so the main point is to check that $\operatorname{WF}(u \otimes v)$ avoids the conormal bundle of the diagonal. The hypothesis excludes exactly the possible conormal directions with opposite covectors, and the zero-covector components in the tensor product estimate are ruled out separately. Finally, the pullback wave front estimate sends a covector pair $(\xi,\eta)$ at $(x,x)$ to $\xi+\eta$, giving the stated inclusion. [/proofplan] [step:Estimate the wave front set of the tensor product distribution] Let $\operatorname{supp} u \subset X$ and $\operatorname{supp} v \subset X$ denote the supports of the distributions $u$ and $v$. The tensor product distribution $u \otimes v \in \mathcal{D}'(X \times X)$ is defined in the standard way by its action on test functions on $X \times X$. By the tensor product wave front set estimate (citing a result not yet in the wiki: Tensor product wave front set estimate), the wave front set of $u \otimes v$ is contained in the subset $S \subset T^*(X \times X)\setminus 0$ defined by \begin{align*} S := S_{11}\cup S_{10}\cup S_{01}, \end{align*} where \begin{align*} S_{11}:=\{(x,y;\xi,\eta):(x,\xi)\in \operatorname{WF}(u),\ (y,\eta)\in \operatorname{WF}(v)\}, \end{align*} \begin{align*} S_{10}:=\{(x,y;\xi,0):(x,\xi)\in \operatorname{WF}(u),\ y\in \operatorname{supp} v\}, \end{align*} and \begin{align*} S_{01}:=\{(x,y;0,\eta):x\in \operatorname{supp} u,\ (y,\eta)\in \operatorname{WF}(v)\}. \end{align*} Thus \begin{align*} \operatorname{WF}(u\otimes v)\subset S. \end{align*} [/step] [step:Identify the conormal bundle of the diagonal] Let $\Delta(X)\subset X\times X$ denote the diagonal submanifold. For $x\in X$, the differential of the diagonal embedding is the [linear map](/page/Linear%20Map) \begin{align*} d\Delta_x:T_xX\to T_{(x,x)}(X\times X),\quad h\mapsto (h,h). \end{align*} Using the natural splitting \begin{align*} T_{(x,x)}^*(X\times X)\cong T_x^*X\oplus T_x^*X, \end{align*} a covector $(\xi,\eta)\in T_x^*X\oplus T_x^*X$ annihilates the tangent space $d\Delta_x(T_xX)$ exactly when \begin{align*} \xi(h)+\eta(h)=0\quad\text{for every }h\in T_xX. \end{align*} This condition is equivalent to $\eta=-\xi$. Therefore the conormal bundle of the diagonal is \begin{align*} N^*\Delta(X)=\{(x,x;\xi,-\xi):x\in X,\ \xi\in T_x^*X\setminus\{0\}\}. \end{align*} [guided] We compute the conormal bundle directly from the definition. The diagonal embedding is the smooth map $\Delta: X\to X\times X$ given by $\Delta(x)=(x,x)$. Its differential at $x$ sends a tangent vector $h\in T_xX$ to the tangent vector $(h,h)\in T_xX\oplus T_xX$: \begin{align*} d\Delta_x(h)=(h,h). \end{align*} A covector at $(x,x)\in X\times X$ can be written, under the natural product identification, as a pair $(\xi,\eta)$ with $\xi,\eta\in T_x^*X$. This covector lies in the conormal space to the diagonal precisely when it annihilates every vector tangent to the diagonal. Hence we require \begin{align*} (\xi,\eta)(d\Delta_x(h))=0\quad\text{for every }h\in T_xX. \end{align*} Substituting $d\Delta_x(h)=(h,h)$ gives \begin{align*} \xi(h)+\eta(h)=0\quad\text{for every }h\in T_xX. \end{align*} This says that the covector $\xi+\eta\in T_x^*X$ is the zero covector, so $\eta=-\xi$. Since conormal directions are taken in the punctured cotangent bundle when testing wave front sets, the zero pair is omitted. Thus \begin{align*} N^*\Delta(X)=\{(x,x;\xi,-\xi):x\in X,\ \xi\in T_x^*X\setminus\{0\}\}. \end{align*} [/guided] [/step] [step:Show that the tensor product wave front set avoids the diagonal conormal bundle] We prove that \begin{align*} S\cap N^*\Delta(X)=\varnothing. \end{align*} Let $(x,x;\zeta,-\zeta)\in N^*\Delta(X)$, where $\zeta\in T_x^*X\setminus\{0\}$. If $(x,x;\zeta,-\zeta)\in S_{11}$, then \begin{align*} (x,\zeta)\in \operatorname{WF}(u)\quad\text{and}\quad (x,-\zeta)\in \operatorname{WF}(v), \end{align*} contradicting the hypothesis. If $(x,x;\zeta,-\zeta)\in S_{10}$, then the second covector component must be $0$, while the conormal form gives the second component $-\zeta\neq 0$, a contradiction. If $(x,x;\zeta,-\zeta)\in S_{01}$, then the first covector component must be $0$, while the conormal form gives the first component $\zeta\neq 0$, again a contradiction. Since $S=S_{11}\cup S_{10}\cup S_{01}$, no point of $S$ lies in $N^*\Delta(X)$. Because $\operatorname{WF}(u\otimes v)\subset S$, we obtain \begin{align*} \operatorname{WF}(u\otimes v)\cap N^*\Delta(X)=\varnothing. \end{align*} [/step] [step:Pull back the tensor product along the diagonal] The pullback theorem for distributions (citing a result not yet in the wiki: Pullback theorem for distributions) applies to the smooth map $\Delta: X\to X\times X$ and the distribution $u\otimes v\in \mathcal{D}'(X\times X)$ because the preceding step proves that $\operatorname{WF}(u\otimes v)$ is disjoint from the conormal bundle $N^*\Delta(X)$. Hence the pullback distribution \begin{align*} \Delta^*(u\otimes v)\in \mathcal{D}'(X) \end{align*} is well-defined. By definition, this distribution is the product $uv$: \begin{align*} uv:=\Delta^*(u\otimes v). \end{align*} [/step] [step:Apply the pullback wave front estimate to obtain the stated inclusion] The pullback wave front estimate gives \begin{align*} \operatorname{WF}(\Delta^*(u\otimes v))\subset \{(x,d\Delta_x^\top(\xi,\eta)):(x,x;\xi,\eta)\in \operatorname{WF}(u\otimes v),\ d\Delta_x^\top(\xi,\eta)\neq 0\}. \end{align*} Here $d\Delta_x^\top:T_x^*X\oplus T_x^*X\to T_x^*X$ is the transpose map characterized by \begin{align*} d\Delta_x^\top(\xi,\eta)(h)=(\xi,\eta)(d\Delta_x(h))\quad\text{for every }h\in T_xX. \end{align*} Since $d\Delta_x(h)=(h,h)$, we have \begin{align*} d\Delta_x^\top(\xi,\eta)=\xi+\eta. \end{align*} Using $\operatorname{WF}(u\otimes v)\subset S$, the three pieces of $S$ give the three pieces of the desired upper bound. From $S_{11}$ we obtain covectors of the form \begin{align*} (x,\xi+\eta)\quad\text{with}\quad (x,\xi)\in \operatorname{WF}(u),\ (x,\eta)\in \operatorname{WF}(v). \end{align*} From $S_{10}$ we obtain covectors of the form $(x,\xi)$ with $(x,\xi)\in \operatorname{WF}(u)$, because $d\Delta_x^\top(\xi,0)=\xi$. From $S_{01}$ we obtain covectors of the form $(x,\eta)$ with $(x,\eta)\in \operatorname{WF}(v)$, because $d\Delta_x^\top(0,\eta)=\eta$. Terms for which $\xi+\eta=0$ contribute no point to the wave front set, since wave front sets lie in the punctured cotangent bundle. Therefore \begin{align*} \operatorname{WF}(uv)\subset \operatorname{WF}(u)\cup \operatorname{WF}(v)\cup \{(x,\xi+\eta)\in T^*X\setminus 0:(x,\xi)\in \operatorname{WF}(u),\ (x,\eta)\in \operatorname{WF}(v)\}. \end{align*} This is the claimed wave front set inclusion, and the proof is complete. [/step]