3 The shear three-point correlation function, and the centers of a triangle

3.1 Definition of the 3PCF

Consider three points $\vec{X}_l$, $1\le l\le 3$, and define their difference vectors $\vec{x}_1=\vec{X}_3-\vec{X}_2$, $\vec{x}_2=\vec{X}_1-\vec{X}_3$, $\vec{x}_3=\vec{X}_2-\vec{X}_1$, so that $\vec{x}_1+\vec{x}_2+\vec{x}_3=\vec{0}$ (see Fig. 1). Each of the three difference vectors will be written as $\vec{x}_l=(x_l \cos \varphi_l,x_l\sin \varphi_l)$, so that $\varphi _l$ is the orientation of the lth side of the triangle. Furthermore, we define $\phi _l$ as being the interior angle of the triangle at the corner $\vec{X}_l$. In order to have general relations, in this paper we will use oriented angles, i.e. we will attach to each angle a sign indicating its orientation. More precisely, we will define $\phi_1\in(-\pi,+\pi)$ to have the same sign of the cross product $\vec{x}_2 \times \vec{x}_3$ (where $\vec{a}\times \vec{b}:=a_1 b_2 -a_2 b_1$), and similarly for $\phi_2$ and $\phi_3$. Note that, since $\vec{x}_1 \times \vec{x}_2 = \vec{x}_2 \times \vec{x}_3 = \vec{x}_3 \times \vec{x}_1$(which follows from the vanishing of the sum of the $\vec{x}_l$, or in a more geometric way, from the fact that each of these cross products equals twice the area of the triangle), all angles $\phi _l$ will be either positive or negative; in particular, they all will be positive if the closed path from $\vec{X}_1$ to $\vec{X}_2$ to $\vec{X}_3$ to $\vec{X}_1$ goes around the triangle counter-clockwise. We also observe that this convention for the angles implies that the sum of the internal angles of the triangle will be $\pm \pi$ depending on the orientation; however, this ambiguity will not generally play a role, since all relations for angles are defined modulo $2 \pi$. In the following we will call a triangle positively (respectively, negatively) oriented if the sum of its internal angle is $+\pi$ ($-\pi$). The relation between the $\phi _l$ and the $\varphi _l$ is given by

 

$$% \varphi_3 - \varphi_2 = {} \pi - \phi_1; \quad \varphi_1 - \varphi_3 = {} \pi - \phi_2; \quad \varphi_2 - \varphi_1 = {} \pi - \phi_3.$$ (1)

Let $\gamma_\mu(\vec{X}_l)$ be the Cartesian components of the shear at point $\vec{X}_l$; they are defined in terms of the deflection potential $\psi(\vec{X})$ by the differential operation $\gamma_1=[\partial^2\psi/(\partial X_1^2)-\partial^2\psi/(\partial X_2^2)]/2$, $\gamma_2= \partial^2\psi/(\partial X_1~\partial X_2)$. We define the Cartesian components of the shear 3PCF as

$$% \gamma_{\mu\nu\lambda}(\vec{x}_1,\vec{x}_2,\vec{x}_3)\equiv \le... ...a_\mu(\vec{X}_1)~\gamma_\nu(\vec{X}_2)~\gamma_\lambda(\vec{X}_3) \right\rangle,$$ (2)

where we have made use of the fact that the shear field is assumed to be a homogeneous random field, so that the 3PCF is invariant under translations; therefore, $\gamma_{\mu\nu\lambda}$ does depend only on the separation vectors $\vec{x}_l$. Although we also assume that the random field is isotropic, the Cartesian components $\gamma_{\mu\nu\lambda}$ of the 3PCF do not depend just on the $x_l=\left\vert \vec{x}_l \right\vert$ (see below).

Figure 1: Definitions of the geometry of a triangle. The $\varphi _l$ are the orientations of the sides of the triangle relative to the positive x1-direction, the $\phi _l$ are the interior angles of the triangle, which are related to the $\varphi _l$ by (1), provided the orientation of the three points is as displayed here.

For any reference direction $\zeta_l$, we can define the tangential and cross components of the shear, $\gamma_{l\rm t}$ and $\gamma_{l\times}$, respectively, at point $\vec{X}_l$ relative to this direction by

$$% \gamma(\vec{X}_l,\zeta_l) \equiv \gamma_{\rm t}(\vec{X}_l,\zeta_l)+{\rm i}\gamma_{\times}(\vec{X}_l,\zeta_l) :$$

$$-\gamma(\vec{X}_l) {\rm e}^{-2{\rm i}\zeta_l} =-\left\lbrack \gam... ...\vec{X}_l)+{\rm i}\gamma_2(\vec{X}_l) \right\rbrack {\rm e}^{-2{\rm i}\zeta_l},$$ = (3)

or explicitly in terms of the components,

 

$$% \gamma_\mu(\vec{X}_l,\zeta_l)=-R_{\mu\nu}(2\zeta_l)\gamma_\nu(\vec{X}_l),$$ (4)

where $R_{\mu\nu}(\varphi)=\delta_{\mu\nu}\cos\varphi+{\epsilon}_{\mu\nu}\sin\varphi$ is the rotation matrix, $\delta_{\mu\nu}$ the Kronecker delta, and ${\epsilon}_{12}=1=-{\epsilon}_{21}$, ${\epsilon}_{11}=0={\epsilon}_{22}$; here and in the following, we identify the tangential component with $\mu=1$, and the cross component with $\mu=2$ when we write transformation equations like (4), and we employ the Einstein summation convention. We notice that, because of the rotation behavior of the shear, we only need to specify the projection direction $\zeta_l$modulo $\pi$.

If the reference directions $\zeta_l$ are defined in terms of the position vectors of the vertices of the triangle, or the side vectors, and thus rotate in the same way as the triangle as whole, then the tangential and cross components of the shear are invariant under rotations and translations of the triangle. For example, the direction $\zeta_l$ could be chosen as the direction $\varphi _l$ of the opposite side of the triangle. The 3PCF of these projected shear components will then depend only on the $\left\vert \vec{x}_l \right\vert$,

 

$$\gamma_{\mu\nu\lambda}^{(\zeta_l)}(x_1,x_2,x_3) = -R_{\mu\alpha}(... ...bda\gamma}(2\zeta_3) \gamma_{\alpha\beta\gamma}(\vec{x}_1,\vec{x}_2,\vec{x}_3).$$ (5)

The orientation of the triangle becomes important if the arguments of the 3PCF are written as the three side lengths; in this case, the triangle is defined only up to a parity transformation. Hence, in all relations involving only the side lengths of a triangle, we will assume that the triangle has positive parity. We shall see in Sect. 4 how the shear 3PCF behaves under parity transformations.

In Eq. (5), the directions $\zeta_l$ are arbitrary. For a given set of three points, i.e., for a given triangle, there are several natural choices for the reference directions; we shall discuss those in the next subsection. One choice was already mentioned above, namely the direction $\varphi _l$ of the side opposite to the corner $\vec{X}_l$. We shall label the corresponding 3PCF with the superscript "s'' (for ``side''),

 

$$\gamma_{\mu\nu\lambda}^{(\rm s)}(x_1,x_2,x_3) = -R_{\mu\alpha}(2\... ...a\gamma}(2\varphi_3) \gamma_{\alpha\beta\gamma}(\vec{x}_1,\vec{x}_2,\vec{x}_3).$$ (6)

3.2 The centers of a triangle, and the corresponding shear projections

In addition to the side projection mentioned above, there are several other natural choices for the directions along which the shear can be projected. For each triangle, one can define a number of "centers''; the direction of the vector connecting the point $\vec{X}_l$ with one of these centers can be used to define convenient components of the 3PCF. The four most important centers are: (1) the centroid of a triangle; it is the point where the side-bisectors intersect; (2) the incenter (center of the incircle), which is the intersection of the three angle-bisectors of the interior angles $\phi _l$; (3) the circumcenter (center of the circumcircle), which is the point of intersection of the three midperpendiculars, and (4) the orthocenter, which is the intersection point of the altitudes.

Figure 2: The orthocenter of a triangle, determined by the intersection of the three altitudes. The angle between the altitude of $\vec{x}_1$ and the side $\vec{x}_3$ is easily obtained from the right triangle formed by this altitude and the two sides $\vec{x}_1$ and $\vec{x}_3$.

We now define the tangential and cross components of the shear relative to the direction of the line connecting the point $\vec{X}_l$with one of these centers. Since the line connecting $\vec{X}_l$ with the orthocenter (the point "H'' in Fig. 2) is perpendicular to the side vector $\vec{x}_l$, we find for the projection of the shear relative to the direction of the orthocenter (labeled with superscript "o'') simply by setting $\zeta_l=\varphi_l+\pi/2$ in (4),

 

$$% \gamma^{(\rm o)}_\mu=-\gamma^{(\rm s)}_\mu.$$ (7)

Therefore, we obtain the shear 3PCF for this projection from Eq. (5)

 

$$% \gamma_{\mu\nu\lambda}^{(\rm o)}(x_1,x_2,x_3)=- \gamma_{\mu\nu\lambda}^{(\rm s)}(x_1,x_2,x_3).$$ (8)

Next we consider the center of the incircle. As this is given by the intersection of the angle-bisectors, we obtain for the direction $\zeta_1$ of the line connecting $\vec{X}_1$ with the incenter (see Fig. 3) $\zeta_1=\varphi_3+\phi_1/2$; hence,

$$\begin{eqnarray*}\gamma_\mu^{(\rm in)}(\vec{X}_1)=-R_{\mu\nu}(2\varphi_3+\phi_1)\gamma_\nu(\vec{X}_1). \end{eqnarray*}$$

Using Eqs. (4) and (7), one finds that the relation between the Cartesian components of the shear and those defined with respect to the ortho-center is

 

$$% \gamma_\mu(\vec{X}_l)=R_{\mu\nu}(-2\varphi_l)\gamma_\nu^{(\rm o)}(\vec{X}_l),$$ (9)

so that

$$\begin{eqnarray*}\gamma_\mu^{(\rm in)}(\vec{X}_1)=-R_{\mu\nu}(2\varphi_3-2\varphi_1+\phi_1)\gamma_\nu^{(\rm o)}(\vec{X}_1). \end{eqnarray*}$$

Since $2\varphi_3-2\varphi_1+\phi_1=-2\pi+2\phi_2+\phi_1=-\pi+\phi_2-\phi_3$, where we used Eq. (1) and the fact that the sum of the $\phi _l$ is $\pi$, one finally obtains

 

$$% \gamma_{\mu\nu\lambda}^{(\rm in)}=R_{\mu\alpha}(\phi_2-\phi_3) ... ...3-\phi_1)R_{\lambda\gamma}(\phi_1-\phi_2) \gamma_{\alpha\beta\gamma}^{(\rm o)},$$ (10)

where the result for the other two points were obtained from the one at $\vec{X}_1$ by cyclic permutation of the indices. It should be noted that in this equation, as in all the following, the angles $\phi _l$are functions of the sidelengths x_l (and the orientation), and not independent geometrical quantities; hence, an equation like (10) contains only the x_l as independent variables.

Figure 3: The incenter of a triangle is given by the intersection of the three interior angle-bisectors.

Next we consider the projection onto the center of the circumcircle. If $\zeta_1$ denotes the direction of the line connecting $\vec{X}_1$ with this center, then the shear components with respect to this center (denoted by the superscript "out'') read

$$% \gamma_\mu^{(\rm out)}=-R_{\mu\nu}(2\zeta_1)\gamma_\nu(\vec{X}_1) =-R_{\mu\nu}(2\zeta_1-2\varphi_1)\gamma_\nu^{(\rm o)}(\vec{X}_1),$$ (11)

where in the second step we used Eq. (9). From Thales' theorem (see Fig. 5) one finds that $\zeta_1-\varphi_3= \pm \pi/2 - \phi_3$, where the sign $\pm$ depends on the orientation of the triangle. Hence, $\zeta_1-\varphi_1= \pm \pi/2 - \phi_3 + \varphi_3 - \varphi_1= \mp\pi/2 + \phi_2 - \phi_3$, and we see that the sign ambiguity does not play a role because of the rotation properties of the shear (cf. comment after Eq. (4)). For the other two points, the corresponding relations are obtained by cyclic permutations, so that

$$% \gamma_{\mu\nu\lambda}^{(\rm out)}=R_{\mu\alpha}(2\phi_2-2\phi_... ...\phi_1)R_{\lambda\gamma}(2\phi_1-2\phi_2) \gamma_{\alpha\beta\gamma}^{(\rm o)}.$$ (12)

Finally, we consider the projection on the center of mass of the triangle, or the centroid (denoted by the superscript "cen''), which is the intersection of the three side-bisectors. Denoting by $\psi_3$the angle between the side bisector of side x₃ and $\vec{x}_3$, we obtain from Fig. 4 that

$$\begin{eqnarray*}\sin\psi_3={x_1\over h_3}\sin\phi_2={x_1 x_2\over h_3 x_3}\sin\phi_3, \end{eqnarray*}$$

where we made use of the sine-theorem, and

$$\begin{eqnarray*}h_3={1\over 2}\sqrt{2x_1^2+2x_2^2-x_3^2} \end{eqnarray*}$$

is the length of the side-bisector of side x₃. From the same figure we also obtain that

$$\begin{eqnarray*}\cos\psi_3={h_3^2+x_3^2/4-x_1^2\over h_3 x_3} ={x_2^2-x_1^2\over 2 h_3 x_3}, \end{eqnarray*}$$

so that

 

$$\cos 2\psi_3 = {\left(x_2^2-x_1^2\right)^2-4 x_1^2x_2^2\sin^2\phi_3\over 4 h_3^2x_3^2};$$

$$\sin 2\psi_3 = {\left(x_2^2-x_1^2\right)x_1x_2\sin\phi_3\over h_3^2x_3^2}\cdot$$ (13)

Analogous relations are obtained for the other two points by cyclic permutations of the indices. The shear components projected toward the centroid are then

$$% \gamma_\mu^{(\rm cen)}(\vec{X}_l)=R_{\mu\nu}(2\psi_l)\gamma^{(\rm s)} =-R_{\mu\nu}(2\psi_l)\gamma^{(\rm o)},$$ (14)

where the components of the rotation matrix can be directly obtained from Eq. (13). For completeness, we give the 3PCF for the centroid,

$$% \gamma_{\mu\nu\lambda}^{(\rm cen)}=-R_{\mu\alpha}(2\psi_1) R_{\nu\beta}(2\psi_2)R_{\lambda\gamma}(2\psi_3) \gamma_{\alpha\beta\gamma}^{(\rm o)}.$$ (15)

Figure 4: The centroid of a triangle is the intersection point of the three side-bisectors.

Figure 5: The circumcenter of a triangle is given by the intersection of the three midperpendiculars. According to Thales' theorem, the side x2 subtends the angle $2\phi _2$ as seen from the circumcenter; from that, the angle between the lines connecting the points $\vec{X}_l$ with the circumcenter are easily obtained by noting that all three of them have equal length.