Normalization Conventions: A Four-Point Worked Example

A connection coefficient is not a number until its source and target bases have been normalized. In CFT language, the same warning applies to a fusing matrix: unit-leading conformal blocks, normalized chiral vertex operators, and a full correlator carry different diagonal factors.

This chapter-closing example makes every factor visible. We choose a nonresonant four-point BPZ problem whose Gauss parameters are $(1/4,3/4,1/2)$. The four gamma quotients reduce to radicals, the solutions admit an elementary trigonometric check, and a diagonal pairing normalization turns the unit-leading connection matrix into a signed Hadamard rotation.

The normalization ladder

$$\begin{gathered} \text{unit-leading Gauss frame} \xrightarrow{\ \times S(z)\ } \text{unit-leading BPZ frame} \\ \xrightarrow{\ \times D_i\ } \text{convention-normalized chiral frame} \xrightarrow{\ \text{pairing and structure data}\ } \text{full correlator}. \end{gathered}$$

The first scalar gauge is shared by both local frames in this example. The generic diagonal matrices $D_i$ need not be shared, so they conjugate the connection matrix as $D_1^{-1}C_{10}D_0$. Later, $N_i:=H_i^{1/2}$ will denote one specific choice: the positive factors that normalize the analytic pairing.

Status of the calculation

Statement	Status
BPZ equation and reduction to the Gauss equation	Exact at finite $b\neq0$
Four unit-leading local solutions	Exact on the declared cut plane
Gamma and radical connection matrices	Exact in the stated row-frame direction
Trigonometric and Wronskian checks	Independent exact checks
Positive diagonal full pairing	Exact analytic completion for real $b>0$
Identification with a physical Liouville correlator	Not claimed without spectrum, three-point, reflection, and antiholomorphic conventions

A reflection-symmetric nonresonant slice

Let

$$Q_{\mathrm L}=b+b^{-1}, \qquad b\neq0,$$

and use centered Liouville momenta $a_{\mathrm L,i}=\alpha_i-Q_{\mathrm L}/2$. Choose

$$\begin{aligned} b\,a_{\mathrm L,0} &= -\frac14, \\ b\,a_{\mathrm L,1} &= \frac14, \\ b\,a_{\mathrm L,\infty} &= \frac14. \end{aligned}$$

In uncentered variables,

$$\begin{aligned} \alpha_0 &= \frac{Q_{\mathrm L}}2 -\frac1{4b}, \\ \alpha_1 = \alpha_\infty &= \frac{Q_{\mathrm L}}2 +\frac1{4b} = Q_{\mathrm L}-\alpha_0. \end{aligned}$$

All three nondegenerate background weights are equal:

$$\Delta_{\mathrm e} := \Delta_0 = \Delta_1 = \Delta_\infty = \frac{Q_{\mathrm L}^2}{4} -\frac1{16b^2}.$$

The light degenerate insertion is $V_{-b/2}$, with

$$\Delta_{\mathrm d} = -\frac12-\frac{3b^2}{4}.$$

The reflected momenta are deliberate. The differential equation sees $a_{\mathrm L,i}^2$, hence only the weights. A normalized Liouville field also remembers the reflection relation between $V_\alpha$ and $V_{Q_{\mathrm L}-\alpha}$, including its nontrivial reflection amplitude.

This slice is nonresonant. Its three oriented exponent differences will be

$$1-C=\frac12, \qquad C-A-B=-\frac12, \qquad A-B=-\frac12.$$

None is an integer, so all three singular points admit ordinary two-column Frobenius frames.

The raw BPZ equation retains both fusion channels

Consider the chiral four-point object

$$\mathscr B(z) = \left\langle V_{\alpha_\infty}(\infty) V_{\alpha_1}(1) V_{-b/2}(z) V_{\alpha_0}(0) \right\rangle_{\mathrm{ch}}.$$

This notation represents either local degenerate fusion branch; the two branches together form the solution space. Null-vector decoupling gives

$$\begin{aligned} 0 ={}& b^{-2}\mathscr B'' - \left( \frac1z+\frac1{z-1} \right)\mathscr B' \\ &+ \left[ \frac{\Delta_{\mathrm e}}{z^2} + \frac{\Delta_{\mathrm e}}{(z-1)^2} - \frac{ \Delta_{\mathrm e}+\Delta_{\mathrm d} }{ z(z-1) } \right]\mathscr B. \end{aligned}$$

The raw local powers follow from the centered-momentum rule

$$\rho_{i,\epsilon} = \frac{bQ_{\mathrm L}}2 +\epsilon b\,a_{\mathrm L,i}, \qquad \epsilon=\pm1.$$

For the chosen momenta they are

Point	$+$ branch	$-$ branch
$z=0$	$(2b^2+1)/4$	$(2b^2+3)/4$
$z=1$	$(2b^2+3)/4$	$(2b^2+1)/4$
$z=\infty$	$\nu_{\infty,+}=-b^2-1/4$	$\nu_{\infty,-}=-b^2-3/4$

At infinity the convention is $\mathscr B(z)\sim z^{-\nu_{\infty,\epsilon}}$. The finite-point branch labels encode the two degenerate fusion rules:

$$\begin{array}{c|cc} & + & - \\ \hline z=0 & \alpha_0-\dfrac b2 & \alpha_0+\dfrac b2 \\ z=1 & \alpha_1-\dfrac b2 & \alpha_1+\dfrac b2 \end{array}.$$

Thus the $+$ label means the minus momentum shift. It does not mean “the larger exponent”: at zero it is the smaller power, while at one it is the larger power.

The raw Fuchs sum is an immediate audit:

$$\begin{aligned} & \rho_{0,+}+\rho_{0,-} + \rho_{1,+}+\rho_{1,-} \\ &\qquad + \nu_{\infty,+}+\nu_{\infty,-} =1. \end{aligned}$$

One scalar gauge gives four unit-leading blocks

Work on

$$\Omega_{01} = \mathbb C \setminus \left( (-\infty,0]\cup[1,\infty) \right),$$

and choose $\Log z$ and $\Log(1-z)$ real on $0<z<1$. Extract the common BPZ gauge

$$\begin{aligned} \mathscr B(z) &= S(z)F(z), \\ S(z) &= z^{(2b^2+1)/4} (1-z)^{(2b^2+3)/4}. \end{aligned}$$

The Gauss parameters from the exact BPZ dictionary are

$$\begin{aligned} A &= \frac12 +b \left( a_{\mathrm L,0} +a_{\mathrm L,1} -a_{\mathrm L,\infty} \right) =\frac14, \\ B &= \frac12 +b \left( a_{\mathrm L,0} +a_{\mathrm L,1} +a_{\mathrm L,\infty} \right) =\frac34, \\ C &= 1+2b\,a_{\mathrm L,0} =\frac12. \end{aligned}$$

Consequently,

$$z(1-z)F'' + \left( \frac12-2z \right)F' - \frac{3}{16}F =0.$$

The unit-leading Gauss bases at zero and one are

$$\begin{aligned} f_0(z) &= {}_2F_1 \left( \frac14,\frac34; \frac12;z \right), \\ g_0(z) &= z^{1/2} {}_2F_1 \left( \frac34,\frac54; \frac32;z \right), \end{aligned}$$

and

$$\begin{aligned} f_1(z) &= {}_2F_1 \left( \frac14,\frac34; \frac32;1-z \right), \\ g_1(z) &= (1-z)^{-1/2} {}_2F_1 \left( \frac14,-\frac14; \frac12;1-z \right). \end{aligned}$$

Define ordered row frames

$$\begin{aligned} \boldsymbol{\mathscr B}_0 &= \left( \mathscr B_{0,+}, \mathscr B_{0,-} \right) = S(z)(f_0,g_0), \\ \boldsymbol{\mathscr B}_1 &= \left( \mathscr B_{1,+}, \mathscr B_{1,-} \right) = S(z)(f_1,g_1). \end{aligned}$$

Because $S(z)/z^{\rho_{0,+}}\to1$ at zero and $S(z)/(1-z)^{\rho_{1,+}}\to1$ at one, the shared gauge is unit-leading at both endpoints. No hidden diagonal matrix enters this particular Gauss-to-BPZ step.

The gamma matrix collapses to radicals

Declare the continuation direction and branch order:

$$\boldsymbol{\mathscr B}_0 = \boldsymbol{\mathscr B}_1C_{10},$$

where rows of $C_{10}$ are the target branches $(+,-)$ at one and columns are the source branches $(+,-)$ at zero. On $\Omega_{01}$, the Gauss connection formula gives

$$\begin{aligned} (C_{10})_{+,+} &= \frac{ \Gamma(\frac12)\Gamma(-\frac12) }{ \Gamma(\frac14)\Gamma(-\frac14) }, \\ (C_{10})_{+,-} &= \frac{ \Gamma(\frac32)\Gamma(-\frac12) }{ \Gamma(\frac34)\Gamma(\frac14) }, \\ (C_{10})_{-,+} &= \frac{ \Gamma(\frac12)^2 }{ \Gamma(\frac14)\Gamma(\frac34) }, \\ (C_{10})_{-,-} &= \frac{ \Gamma(\frac32)\Gamma(\frac12) }{ \Gamma(\frac34)\Gamma(\frac54) }. \end{aligned}$$

Use

$$\begin{gathered} \Gamma\left(-\frac12\right) =-2\sqrt\pi, \qquad \Gamma\left(\frac32\right) =\frac{\sqrt\pi}{2}, \\ \Gamma\left(-\frac14\right) =-4\Gamma\left(\frac34\right), \qquad \Gamma\left(\frac54\right) =\frac14\Gamma\left(\frac14\right), \\ \Gamma\left(\frac14\right) \Gamma\left(\frac34\right) =\pi\sqrt2. \end{gathered}$$

Every special-function constant cancels:

$$C_{10} = \begin{pmatrix} \dfrac1{2\sqrt2} & -\dfrac1{\sqrt2} \\ \dfrac1{\sqrt2} & \sqrt2 \end{pmatrix}, \qquad \det C_{10}=1.$$

Equivalently,

$$\begin{aligned} f_0 &= \frac1{2\sqrt2}f_1 + \frac1{\sqrt2}g_1, \\ g_0 &= -\frac1{\sqrt2}f_1 + \sqrt2\,g_1. \end{aligned}$$

The inverse direction is

$$C_{01} = C_{10}^{-1} = \begin{pmatrix} \sqrt2 & \dfrac1{\sqrt2} \\ -\dfrac1{\sqrt2} & \dfrac1{2\sqrt2} \end{pmatrix}.$$

Transposing $C_{10}$ would describe the scalar-coefficient convention with source labels on rows. It would not describe the declared right-acting row-frame relation.

A trigonometric uniformization checks every entry

On the real interval $0<z<1$, set

$$z=\sin^2\vartheta, \qquad \varphi=\frac\pi2-\vartheta, \qquad 0<\vartheta<\frac\pi2.$$

The four Gauss functions reduce to

$$\begin{aligned} f_0 &= \frac{\cos(\vartheta/2)}{\cos\vartheta}, & g_0 &= \frac{2\sin(\vartheta/2)}{\cos\vartheta}, \\ f_1 &= \frac{2\sin(\varphi/2)}{\sin\varphi}, & g_1 &= \frac{\cos(\varphi/2)}{\sin\varphi}. \end{aligned}$$

Since

$$\sin\varphi=\cos\vartheta, \qquad \frac{\varphi}{2} = \frac\pi4-\frac{\vartheta}{2},$$

the two connection relations become the addition identities

$$\begin{aligned} \cos\frac{\vartheta}{2} &= \frac1{\sqrt2} \left( \sin\frac{\varphi}{2} + \cos\frac{\varphi}{2} \right), \\ 2\sin\frac{\vartheta}{2} &= \sqrt2 \left( \cos\frac{\varphi}{2} - \sin\frac{\varphi}{2} \right). \end{aligned}$$

This verifies all four radical coefficients without gamma algebra. Analytic continuation then extends the equality from the interval to the chosen cut plane.

Wronskians fix the determinant before simplification

For the Gauss frames, Abel’s identity and the unit-leading endpoint coefficients give

$$\Wr[f_0,g_0] = \frac12 z^{-1/2} (1-z)^{-3/2},$$

and

$$\Wr[f_1,g_1] = \frac12 z^{-1/2} (1-z)^{-3/2}.$$

Because $(f_0,g_0)=(f_1,g_1)C_{10}$,

$$\Wr[f_0,g_0] = \det(C_{10}) \Wr[f_1,g_1].$$

The Wronskians therefore force $\det C_{10}=1$ before any gamma function is evaluated. This agrees with the generic centered-momentum audit

$$\det C_{10} = -\frac{a_{\mathrm L,0}}{a_{\mathrm L,1}} =1.$$

Multiplication by the shared scalar gauge multiplies each Wronskian by $S(z)^2$. Hence the raw BPZ frames obey

$$\begin{aligned} \Wr[ \mathscr B_{0,+}, \mathscr B_{0,-} ] &= \Wr[ \mathscr B_{1,+}, \mathscr B_{1,-} ] \\ &= \frac12 \left[ z(1-z) \right]^{b^2}. \end{aligned}$$

This is also the direct Abel solution of the specialized BPZ equation. It checks the raw powers, scalar gauge, basis order, and matrix direction at once.

Pairing normalization turns the matrix into a rotation

The unit-leading matrix is not orthogonal. That is not a defect: orthogonality refers to a chosen pairing, not to Frobenius leading coefficients.

For real $b>0$, pair antiholomorphic blocks by complex conjugation and choose, up to an overall positive scalar,

$$H_0 = \begin{pmatrix} 1&0 \\ 0&\frac14 \end{pmatrix}.$$

The connection relation requires

$$\begin{aligned} H_1 &= C_{10}H_0C_{10}^{\mathsf T} \\ &= \begin{pmatrix} \frac14&0 \\ 0&1 \end{pmatrix}. \end{aligned}$$

Both matrices are diagonal in their local channel. The resulting monodromy-invariant analytic completion can be written in either frame:

$$\begin{aligned} & |\mathscr B_{0,+}|^2 + \frac14|\mathscr B_{0,-}|^2 \\ &\qquad = \frac14|\mathscr B_{1,+}|^2 + |\mathscr B_{1,-}|^2. \end{aligned}$$

Now take the positive square roots

$$N_0 = \operatorname{diag} \left( 1,\frac12 \right), \qquad N_1 = \operatorname{diag} \left( \frac12,1 \right),$$

and define pairing-normalized frames

$$\widehat{\boldsymbol{\mathscr B}}_i = \boldsymbol{\mathscr B}_iN_i.$$

Their connection matrix is

$$\begin{aligned} \widehat C_{10} &= N_1^{-1}C_{10}N_0 \\ &= \frac1{\sqrt2} \begin{pmatrix} 1&-1 \\ 1&1 \end{pmatrix}. \end{aligned}$$

Thus

$$\widehat C_{10} \widehat C_{10}^{\mathsf T} =I, \qquad \det\widehat C_{10}=1.$$

The nonorthogonal radical matrix and the orthogonal signed-Hadamard rotation encode the same analytic continuation map in two different normalizations.

We therefore call $\widehat C_{10}$ a pairing-normalized connection matrix. It becomes a particular degenerate CFT fusing matrix only if the factors $N_i=H_i^{1/2}$ are also adopted as the chiral-vertex normalization; a physical theory need not make that choice.

The unit-leading matrix has $\det C_{10}=1$ but is not orthogonal. Diagonal rescalings by $N_0$ and $N_1$ preserve the continuation map and produce the pairing-normalized rotation $\widehat C_{10}=N_1^{-1}C_{10}N_0$.

General diagonal covariance

For arbitrary invertible diagonal matrices $D_i$, set

$$\widetilde{\boldsymbol{\mathscr B}}_i = \boldsymbol{\mathscr B}_iD_i.$$

Then

$$\widetilde C_{10} = D_1^{-1}C_{10}D_0,$$

while a fixed full pairing is represented by

$$\widetilde H_i = D_i^{-1} H_i \left( D_i^{-1} \right)^\dagger.$$

Its determinant changes as

$$\det\widetilde C_{10} = \frac{\det D_0}{\det D_1} \det C_{10}.$$

Therefore neither the individual entries nor the determinant of a connection matrix are invariant under independent source and target normalizations. Monodromy conjugacy classes, a fully normalized pairing, or a complete boundary observable are the appropriate invariant objects.

What this example does—and does not—normalize

The acronym DOZZ denotes the Dorn–Otto–Zamolodchikov–Zamolodchikov three-point structure constants. The complete ledger is:

Layer	Data fixed here	Remaining convention or input
Gauss ODE	Unit-leading functions and cut plane	None within the declared branch domain
Raw BPZ block	Shared scalar gauge $S(z)$, branch order, and inherited field-at-infinity convention	None within the declared CFT convention
Chiral vertex block	Covariance law $\widetilde C_{10}=D_1^{-1}C_{10}D_0$	Actual diagonal $D_i$ and three-point chiral-vertex normalization
Analytic full pairing	Complex conjugation and the relative matrices $H_0,H_1$	One common positive scalar
Physical Liouville correlator	Not fixed	Spectrum, DOZZ coefficients, reflection amplitude, and integration or summation prescription

In particular, the equality $\alpha_1=Q_{\mathrm L}-\alpha_0$ does not identify the corresponding normalized Liouville fields with coefficient one. The ODE cannot recover that reflection coefficient because it depends only on the equal weights.

The example also avoids resonance by design. If $C$ or $C-A-B$ becomes an integer, the displayed unit-leading pair degenerates and gamma poles must be combined with a singular basis change before taking the limit. The finite result is a Frobenius–logarithmic frame. A pole of a gamma factor by itself is not evidence that the solution space diverges, and a reciprocal-gamma zero can instead signal truncation or reducibility.

Common pitfalls

Quoting the matrix without its direction. Here $\boldsymbol{\mathscr B}_0=\boldsymbol{\mathscr B}_1C_{10}$. Changing to column frames, scalar coefficient arrays, or the inverse continuation changes the displayed matrix.

Forgetting the branch order. Rows are the $(+,-)$ branches at one; columns are the $(+,-)$ branches at zero. The $+$ sign labels the momentum shift $\alpha_i-b/2$, not a globally larger exponent.

Calling the unit-leading matrix a normalized fusing matrix. It is a perfectly normalized ODE connection matrix. A chosen CFT chiral-vertex normalization can still multiply its rows and columns diagonally.

Demanding orthogonality before choosing a pairing. Frobenius normalization fixes leading coefficients, not inner products. Orthogonality appears only after the explicit $H_i$ normalization.

Treating reflection as literal equality of fields. Reflected momenta have equal conformal weights, so they give the same BPZ coefficient. Their physical Liouville fields differ by a normalization-dependent reflection amplitude.

Substituting resonant parameters into the generic gamma matrix. Gamma poles then diagnose a degenerating basis. Construct the logarithmic basis first and only then take the limit.

Exercises

1. Recover the Gauss parameters

Insert the chosen centered momenta into the BPZ-to-Gauss dictionary and derive $(A,B,C)=(1/4,3/4,1/2)$.

Solution

The three combinations are

$$\begin{aligned} b \left( a_{\mathrm L,0} +a_{\mathrm L,1} -a_{\mathrm L,\infty} \right) &=-\frac14, \\ b \left( a_{\mathrm L,0} +a_{\mathrm L,1} +a_{\mathrm L,\infty} \right) &=\frac14, \\ 2b\,a_{\mathrm L,0} &=-\frac12. \end{aligned}$$

Adding the constant terms in the dictionary gives $A=1/4$, $B=3/4$, and $C=1/2$.

2. Audit the raw powers

Derive the six raw powers and verify the Fuchs relation.

Solution

Since $bQ_{\mathrm L}/2=(b^2+1)/2$, substitution into $\rho_{i,\epsilon}=bQ_{\mathrm L}/2+\epsilon b a_{\mathrm L,i}$ gives the two finite-point rows in the table. At infinity,

$$\nu_{\infty,\epsilon} = -\frac12-b^2 +\epsilon b\,a_{\mathrm L,\infty},$$

which gives $-b^2-1/4$ and $-b^2-3/4$. The sum is

$$(b^2+1)+(b^2+1)+(-2b^2-1)=1.$$

3. Reduce the gamma matrix

Use recurrence and reflection identities to reduce all four gamma quotients to radicals.

Solution

The recurrence formula gives

$$\Gamma\left(-\frac12\right) =-2\sqrt\pi, \quad \Gamma\left(\frac32\right) =\frac{\sqrt\pi}{2},$$

and

$$\Gamma\left(-\frac14\right) =-4\Gamma\left(\frac34\right), \quad \Gamma\left(\frac54\right) =\frac14\Gamma\left(\frac14\right).$$

Euler reflection gives $\Gamma(1/4)\Gamma(3/4)=\pi\sqrt2$. Entrywise substitution yields

$$C_{10} = \begin{pmatrix} \dfrac1{2\sqrt2}&-\dfrac1{\sqrt2} \\ \dfrac1{\sqrt2}&\sqrt2 \end{pmatrix}.$$

4. Prove the connection formula without gamma functions

Set $z=\sin^2\vartheta$ and use the four trigonometric forms to recover both columns of $C_{10}$.

Solution

With $\varphi/2=\pi/4-\vartheta/2$,

$$\begin{aligned} \sin\frac{\varphi}{2} +\cos\frac{\varphi}{2} &= \sqrt2\cos\frac{\vartheta}{2}, \\ \cos\frac{\varphi}{2} -\sin\frac{\varphi}{2} &= \sqrt2\sin\frac{\vartheta}{2}. \end{aligned}$$

Divide by $\sin\varphi=\cos\vartheta$ and insert the numerical coefficients. The first identity gives the expansion of $f_0$; the second gives the expansion of $g_0$.

5. Obtain the determinant from Wronskians

Show directly that the unit-leading zero and one frames have the same Wronskian.

Solution

For the Gauss equation,

$$\Wr' = - \frac{ C-(A+B+1)z }{ z(1-z) } \Wr.$$

With $(A,B,C)=(1/4,3/4,1/2)$, integration gives

$$\Wr = Kz^{-1/2}(1-z)^{-3/2}.$$

The unit-leading local expansions at either endpoint give $K=1/2$. Therefore the two Wronskians agree, and the declared frame relation forces $\det C_{10}=1$.

6. Solve for the diagonal pairing

Let $H_0=\operatorname{diag}(1,r)$. Require $C_{10}H_0C_{10}^{\mathsf T}$ to be diagonal. Find $r$ and the pairing-normalized connection matrix.

Solution

Write

$$C_{10} = \begin{pmatrix} a&-c \\ c&d \end{pmatrix}.$$

The off-diagonal entry of $C_{10}H_0C_{10}^{\mathsf T}$ is $c(a-rd)$. Thus

$$r=\frac ad=\frac14.$$

The transformed pairing is $H_1=\operatorname{diag}(1/4,1)$. Its positive square roots give

$$\widehat C_{10} = \frac1{\sqrt2} \begin{pmatrix} 1&-1 \\ 1&1 \end{pmatrix}.$$

7. Track an arbitrary diagonal normalization

Suppose $D_0=\operatorname{diag}(u_+,u_-)$ and $D_1=\operatorname{diag}(v_+,v_-)$. Determine how the connection matrix and its determinant change.

Solution

From $\widetilde{\boldsymbol{\mathscr B}}_i =\boldsymbol{\mathscr B}_iD_i$,

$$\widetilde C_{10} = D_1^{-1}C_{10}D_0.$$

Consequently,

$$\det\widetilde C_{10} = \frac{u_+u_-}{v_+v_-} \det C_{10}.$$

The entries and determinant therefore depend on the two local normalizations even though the underlying continuation map does not.

References

• A. A. Belavin, A. M. Polyakov, and A. B. Zamolodchikov, “Infinite Conformal Symmetry in Two-Dimensional Quantum Field Theory”, Nuclear Physics B 241 (1984), 333–380. Equations (5.17)–(5.24) derive the second-order null-vector equation, its four-point Riemann form, and the two fusion branches.
• G. Bonelli, C. Iossa, D. Panea Lichtig, and A. Tanzini, “Irregular Liouville Correlators and Connection Formulae for Heun Functions”, Communications in Mathematical Physics 397 (2023), 635–727. Equations (2.1.1)–(2.1.7) give the centered-momentum BPZ equation, hypergeometric solutions, full pairing, and gamma matrix used here.
• J. Teschner, “Liouville Theory Revisited”, Classical and Quantum Gravity 18 (2001), R153–R222. Equations (209), (222), (224), and (231)–(252) distinguish chiral-vertex normalizations, fusion transformations, and full-correlator crossing.
• NIST Digital Library of Mathematical Functions, §15.10, especially equations 15.10.1–15.10.5 and 15.10.21–15.10.22, for the Gauss equation, canonical bases, Wronskians, and connection coefficients.
• NIST Digital Library of Mathematical Functions, §15.4, especially equations 15.4.12, 15.4.14, 15.4.16, and 15.4.18, for the elementary trigonometric reductions used in the independent check.