In the following series of notes, we want to understand the famous paper [KT01] by H. Koch and D. Tataru. We first fix some notations. We denote by $\hat\cdot$ or $\mathcal{F}(\cdot)$ the Fourier transform, $\cdot^t$ the transpose of matrix (or vector), and $i$ the imaginary unit (i.e. $i^2=-1$). Here the domain considered is the whole space $\mathbb{R}^n$. We begin with the Leray projection operator, which is also called the divergence free operator. === Introduction to the Leray projection operator $\Pi$. $\Pi$ is defined by matrix valued Fourier multiplier $m(\xi)$ where $$m(\xi) = \delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2},$$ which is bounded, and satisfies the Mihlin-Hormander condition $$\sup_{\xi\neq 0} \abs{\xi}^{\abs{\alpha}} \abs{\p_\xi^\alpha m(\xi)} \leq C_\alpha,\hspace{6pt} \forall\, \alpha.$$ Formally, $\forall\, v=(v_1,\cdots,v_n)^t \in \mathcal{S}(\mathbb{R}^n)^n$, we have $$(\Pi v)_k = \f{1}{\sqrt{2\pi}^n }\int (\delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2})\hat v_j e^{ix\cdot\xi} d\xi,$$ where we used the Einstein summation convention. The Leray projection operator $\Pi$ commutes with the Laplacian operator $\Delta$ (with Fourier multiplier $-\abs{\xi}^2$) since the operators induced by multiplier commute with each other. **Claim 1:** $\grad\cdot\Pi v = 0$, for all $v\in \mathcal{S}(\mathbb{R}^n)^n$. (This is why $\Pi$ is called Divergence free operator) $$\begin{split} \grad\cdot\Pi v=\sum_{k} \p_{x_k}(\Pi v)_k&=\f{1}{\sqrt{2\pi}^n }\int (\delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2})\hat v_j i\xi_k e^{ix\cdot\xi} d\xi~\\ &=\f{1}{\sqrt{2\pi}^n }\int \sum_k(\delta_{kj}\xi_k - \f{\xi_k^2\xi_j}{\abs{\xi}^2})\hat v_j i e^{ix\cdot\xi} d\xi~\\ &=0. \end{split}$$ **Claim 2:** $\Pi\grad f=0$, for all $f\in \mathcal{S}(\mathbb{R}^n)$. Note that $\widehat{\grad f} = (i\xi_1\hat f,\cdots,i\xi_n\hat f)^t$, hence $$\begin{split} \Pi\grad f &= \f{1}{\sqrt{2\pi}^n }\int (\delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2})i\xi_j\hat f e^{ix\cdot\xi} d\xi~\\ &=\f{1}{\sqrt{2\pi}^n }\int (\delta_{kj}\xi_j - \f{\xi_k\xi_j^2}{\abs{\xi}^2})i\hat f e^{ix\cdot\xi} d\xi~\\ &=0. \end{split}$$ **Remark 1** The main calculations for **Claim 1** and **Claim 2** are exactly the same. **Claim 3** $\Pi(\Pi v) = \Pi v$. This is why $\Pi$ is called Projection operator. By definition, we have $$(\widehat{\Pi v})_k = (\delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2})\hat v_j.$$ Hence $$\begin{split} (\widehat{\Pi(\Pi v)})_l &=(\delta_{lk} - \f{\xi_l\xi_k}{\abs{\xi}^2})(\widehat{\Pi v})_k ~\\ &=(\delta_{lk} - \f{\xi_l\xi_k}{\abs{\xi}^2}) (\delta_{kj} - \f{\xi_k\xi_j}{\abs{\xi}^2})\hat v_j ~\\ &=(\delta_{lk}\delta_{kj} - \delta_{kj}\f{\xi_l\xi_k}{\abs{\xi}^2} - \delta_{lk} \f{\xi_k\xi_j}{\abs{\xi}^2} + \f{\xi_l\xi_k^2\xi_j}{\abs{\xi}^4})\hat v_j ~\\ &=(\text{taking summation first on k}) ~\\ &=(\delta_{lj} - \f{\xi_l\xi_j}{\abs{\xi}^2})\hat v_j = (\widehat{\Pi v})_l, \end{split}$$ which establishes the claim. === Alternative definition of the operator $\Pi$. Let $R_j$ be the Riesz transform with Fourier multiplier $\f{\xi_j}{\abs{\xi}}$, then $$(\Pi v)_k = v_k - R_k \sigma,$$ where $\sigma = \sum_j R_j v_j$. === References [KT01] H. Koch and D. Tataru, Well-posedness for the Navier-Stokes equations, Adv. Math. **157**, 22-35 (2001)