# Wave front set of vector-valued Dirac delta distribution

Context: I am reading a physics paper Local Wick Polynomials and Time Ordered Products of Quantum Fields in Curved Spacetime which applies the notion of the wave front set to operator-valued distributions in a globally hyperbolic spacetime. In particular, the paper considers vector-valued distributions $W_n$, whose wave front set is (eq. 9)

$$\text{WF}(W_n)=\left\{ (x_1,k_1,\dots,x_n,k_n)\in(T^*M)^n \backslash \left\{0\right\}|k_i\in V^-_{x_i},i=1,\dots,n \right\},$$
where $V^\pm_{x_i}$ denotes the future/past lightcone in the tangent space of $x_i\in M$. It is then claimed that we will get well-defined products between these $W_n$ distributions and distributions whose wave front sets are subsets of

$$G_n(M,g)\equiv (T^* M)^n \backslash \left(\bigcup_{x\in M} (V_x^+)^n\cup\bigcup_{x\in M} (V_x^-)^n \right),$$ where $g$ denotes the spacetime metric.

It is then asserted that the wave front set of the distribution $$t(x_1,\dots,x_n)\equiv f(x_1)\delta (x_1,\dots,x_n) \tag{eq. 16} \label{t}$$ is a subset of $G_n(M,g)$ because its wave front set is $$\text{WF}(t)=\left\{(x,k_1,\dots,x,k_k)\in (T^*M)^k \backslash \left\{0\right\}|\sum_i k_i =0 \right\}. \tag{*} \label{WFt}$$

Questions:

1) It is not clear to me why the wave front set of \eqref{t} is given by \eqref{WFt}. More descriptively, taking the Fourier transform of, e.g., $\delta(x_1,x_2)=\delta (x_1)\otimes \delta (x_2)$, I get $$\delta(e^{ix_1\cdot k_1},e^{ix_2\cdot k_2})=(e^{i0\cdot k_1},e^{i0\cdot k_2})=(1,1),$$ which suggests to me that the wave front set of \eqref{WFt} is instead $(0,k_1,\dots,0,k_n)$ for $k_i\in \mathbb{R}^n$

2) Even if I were accept that the wave front set of \eqref{t} is given by \eqref{WFt}, it is not clear to me how the condition $\sum_i k_i=0$ implies that this wave front set is a subset of $G_n(M,g)$. In particular, suppose $i=2$ and thus $k_1+k_2=0 \rightarrow k_1=-k_2$. Now, if $k_1\in{V^+_x}$, then $k_2\in{V^-_x}$ which would seem to imply that this wave front set is not a subset of $G_n(M,g)$.

• Not really clear why the question has a down vote. Is it because it contains the words "physics paper"? Commented May 2, 2017 at 19:25

(1) A careful reading of the paper will reveal that the notation $\delta(x_1,\ldots,x_n)$ is precisely the $\delta$-distribution supported on the diagonal of the $n$-fold Cartesian product $M \times \cdots \times M$. If we look at (a possibly small portion of) the diagonal as covered by the coordinates $(x_1,\ldots,x_n)$, then $\delta(x_1,\ldots,x_n) = \delta(x_2-x_1) \cdots \delta(x_n-x_1)$, or any other equivalent form, rather than $\delta(x_1-0)\cdots \delta(x_n-0)$. It is not hard to see that, on $(\mathbb{R}^{\dim M})^n$, the Fourier transform of the $\delta$-distribution supported on the diagonal is $\delta(\sum_{i=1}^n k_i)$. From this you can read off that for a general manifold $M$ has the form (*), $$\mathrm{WF}(\delta(x_1,\ldots,x_n)) = \{ (x,k_1,\ldots, x,k_n) \in (T^*M)^n \setminus \{0\} \mid x\in M, \sum_{i=1}^n k_i = 0 \} .$$
(2) Over the diagonal of $M^n$, a non-zero element of $(T^*M)^n \setminus G_n(M,g)$ is of the form $(x,k_1,\ldots, x,k_n)$ with either all $k_i$ in $V_x^+$ or all $k_i$ in $V_x^-$. In particular, by the convexity of the cones, either $\sum_{i=1}^n k_i$ in $V_x^+$ or in $V_x^-$, meaning that this sum could not be zero. In other words, the condition $\sum_{i=1}^n k_i = 0$ forces $\mathrm{WF}(\delta(x_1,\cdots,x_n))$ to be a subset of $G_n(M,g)$, as claimed in the paper.
• @user143410, please read carefully: it is the complement of $G_n(M,g)$ that confines all $k_i$ to either the future or the past lightcone, as should be clear from its definition. Commented May 2, 2017 at 8:11
• @user143410 Unfortunately, I'm having trouble understanding your confusion. The set $G_n(M,g)$ is defined in eq. (11) of the original paper and reproduced correctly in your question. The formula in the image that you link to is different and I don't see a reason for it to enter the discussion. If the question is to clarify the notation in the definition of $G_n(M,g)$, let me simply restate it in words: Take $(T^*M)^n \cong T^*(M^n)$ and, for every $x\in M$,remove every point of the form $(x,k_1,\ldots,x,k_n)$ where all the $k_i$ are simultaneously in $V_x^+$ or simultaneously in $V_x^-$. Commented May 2, 2017 at 15:35
• @user143410 Ah, now I understand what you are asking. And the answer is simply yes, the last version of your formula, correctly written as $(T^*M)^n \setminus \left(\bigcup_{x\in M} (V_x^+ \cup V_x^-)^n \right)$, is correct. Commented May 2, 2017 at 19:23