TITLE: Write $x^k - y^k$ as the product of two factors.
QUESTION [3 upvotes]: Write $x^k - y^k$ as the product of two factors.
$x^3 - y^3 = (x-y)(x^2+xy+y^2)$
$x^4 - y^4 = (x-y)(x^3+x^2y+xy^2+y^3)$
$x^5 - y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
$x^6 - y^6 = (x-y)(x^5+x^4y+x^3y^2+x^2y^3+xy^4+y^5)$
$x^k - y^k = (x-y)(x^{k-1}+\underbrace{x^{k-2}y + \cdots + xy^{k-2}}_{k-2} +y^{k-1})$
It's not exact what the middle terms should be, so I'm not sure how to write the expression for the second factor. I think I may need to use the parity of $k$ and some mention of mirroring, though mirroring sounds wrong.

REPLY [2 votes]: Here is a variation without using induction. We want to use the sigma symbol and start with an example for small $n=4$ to better see what's going on.
We have
\begin{align*}
x^4-y^4&=(x-y)(x^3+x^2y+xy^2+y^3)\\
&=(x-y)(x^3y^0+x^2y^1+x^1y^2+x^0y^3)\tag{1}\\
&=(x-y)\sum_{k=0}^3 x^{3-k}y^k
\end{align*}
Since we can write $x^3=x^3y^0$ and $y^3=x^0y^3$ each term in (1) has a representation $x^{3-k}y^k$, $0\leq k\leq 3$ which can be conveniently written using the sigma symbol $\sum$.

For the general case with integral $n>0$ we have
\begin{align*}
\color{blue}{x^n-y^n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots + xy^{n-2}+y^{n-1})\\
&=(x-y)(x^{n-1}y^0+x^{n-2}y^1+\cdots + x^1y^{n-2}+x^0y^{n-1})\\
&\,\,\color{blue}{=(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}\tag{2}
\end{align*}

We prove (2) algebraically. We obtain for integral $n>0$ by starting with the right-hand side of  (2)
\begin{align*}
\color{blue}{(x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k}&=x\sum_{k=0}^{n-1} x^{n-1-k}y^k-y\sum_{k=0}^{n-1} x^{n-1-k}y^k\tag{3.1}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=0}^{n-1} x^{n-1-k}y^{k+1}\tag{3.2}\\
&=\sum_{k=0}^{n-1} x^{n-k}y^k-\sum_{k=1}^{n} x^{n-k}y^{k}\tag{3.3}\\
&\,\,\color{blue}{=x^n-y^n}
\end{align*}
and the claim follows.
Comment:

In (3.1) we multiply out $(x-y)\sum \Box=x\sum \Box - y\sum\Box$.

In (3.2) we simplify by multiplying with $x$ resp. $y$. Note, we have $x\sum x^{n-1-k}y^k=\sum x^{1+(n-1-k)}y^k=\sum x^{n-k}y^k$.

In (3.3) we shift the index $k$ in the right-hand side by one. We so have lower bound $k=1$ and upper bound $k=n$. We compensate this shift by substituting $k\to k-1$ within the sum. We now have the convenient situation that we have a telescoping sum with alike terms in the left-hand sum and the right-hand sum for $k=1,\ldots,n-1$, so that everything cancels out besides $x^n-y^n$.


TITLE: Prove that the ring of rational numbers $\Bbb Q$ is not isomorphic to the ring of real numbers $\Bbb R$
QUESTION [3 upvotes]: just wondering if my reasoning is correct.
I said assume there is such a homomorphism f, then f(1)=1 since it is a ring homomorphism. But
$$f(\sqrt 2)= f(1\cdot\sqrt 2)= f(1) \cdot \sqrt 2= \sqrt 2$$
but $\sqrt 2$ is not a rational, contradiction. Is the proof this simple or am i missing something? Thanks in advance.

REPLY [7 votes]: No, that reasoning is not correct, but it is almost correct. $\sqrt{2}$ doesn't mean anything in a general ring.
I will subscript things with $_q$ for rationals an $_r$ for reals when something is ambiguous. So $2_q$ is the rational number $2$, while $2_r$ is the real number.
$$2_q=1_q+1_q=f(1_r)+f(1_r)=f(2_r)=f(\sqrt{2}\cdot \sqrt{2})=f(\sqrt{2})\cdot f(\sqrt{2})$$
So $f(\sqrt{2})$ would have to be a rational number whose square is $2$, which is impossible.
This is not the same as saying $f(\sqrt{2})=\sqrt{f(2)}$, because square root is not a function on $\mathbb Q$, so it doesn't actually mean anything.


TITLE: Why is the Petersson inner product positive definite?
QUESTION [2 upvotes]: The Petersson inner product is defined on the space $\mathcal{S}_k(\Gamma)$ of weight $k$ cusp forms of level $\Gamma$, and takes values in $\mathbb{C}$.
First of all, I wonder: what does it mean for a complex-valued inner product to be positive definite?
Then, can anyone show me why the Petersson inner product is positive definite?

REPLY [2 votes]: what does it mean for a complex-valued inner product to be positive definite?

A Hermitian form $\langle \cdot,\cdot\rangle $ is positive definite if $\langle f,f\rangle >0$ whenever $f\ne 0$.

why the Petersson inner product is positive definite

Because plugging $g=f$ in the definition, we get an integral of a nonnegative function (which is zero a.e. if and only if $f=0$ a.e.). One should worry a little about the integral being convergent and appropriately behaved under the action of the modular group, but this is really the issue of the product being well-defined rather than positive-definite. You can find the details on page 74-75 of Introduction to the Arithmetic Theory of Automorphic Functions by Shimura.


TITLE: Limit definition applications
QUESTION [1 upvotes]: Is there a way to use the limit definition to prove that the limit as $x$ approaches 0 of $\dfrac{\sin x} x = 1$? I get the definition (sort of), but it doesn't seem to have any practical applications in terms of extension of knowledge, just raw verification and mathematical rigor.
Thanks.
EDIT: John: it was a typo. Sorry for the confusion, the limit of sin(x)/x as x approaches infinity seems much more obvious. I am more interested as x approaches 0. My question really comes down to a much simpler concept: it seems like proofs often reveal to us new information through a series of logically consequent statements. However, I can't tell what the new information gained by epsilon delta proofs are, and I was wondering if there was some new mathematical idea that was revealed by this method.

REPLY [0 votes]: This is very incorrect
Given
and
$$ f(x) = \frac{-1}{x}$$
$$ h(x) = \frac{1}{x} $$
since sin(x) is alway between range[-1, 1]
$$ f(x) \leq sin(x) \geq h(x)$$
concluding that for all x in
$$(x, 0)$$
$$ \lim_{n \to 0} f(x) = 0$$
and
$$ \lim_{n \to 0} g(x) = 0$$
Concluding that
$$ \lim_{x \to 0} \frac{sin(x)}{x}= 0 $$
You can aslo try both negative and positive limits to find your answer.
because we have proven that the limit as x approaches negative infinity or positive infinity is zero or that this limit doesn't exist.
Conclusion
since $$0\leq\sin(X)\geq0$$
We conclude the limit of sin(x) as x approaches 0 must be 0


TITLE: Lie derivative isn't $\lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}$ in $\mathbb{R^n}$? Why?
QUESTION [2 upvotes]: The Lie derivative o a vector field $W$ along another vector field $V$ at point p (denoted by $(L_V W)_p$) is defined as the limit when $t \rightarrow 0$ of $\frac{(\theta_{-t})_*W_{\theta_t(p)}-W_p}{t}$ where $(-)_*$ denotes the pushforward and $\theta_t(p)$ the flow of $V$. Suppose we are in $\mathbb{R^n}$. For me, it seems natural that in this case, the Lie derivative would be:
$$(L_V W)_p=\lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}.$$
But suppose I take as example $V(x,y)=(3x,4y)$. Then the flow, if we denote a point $p=(x,y)$ is given by $\theta_t(p)=(xe^{3t},ye^{4t})$ so the the pushforward (which as I understand is the Jacobian matrix here) is:
$$D (\theta_{-t}) =
\begin{bmatrix}
e^{-3t} && 0 \\
0 && e^{-4t}
\end{bmatrix}$$
But then we would have $(L_V W)_p=\lim_{t\rightarrow 0} \frac{D (\theta_{-t})W_{\theta_t(p)}-W_p}{t} \neq \lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}$ which is not so intuitive, no?

REPLY [1 votes]: It is not intuitive if you think about all tangent spaces being "fixed". For example, take the curve $\gamma:t\mapsto(\cos t,\sin t)$. Its tangent vector is given by $V=(-\sin t,\cos t)$.
Imagine that you look above the $\mathbb{R}^2$ plane while $t$ grows: what you see is an arrow, based on points of a circle and tangent to this same circle, making turns around the origin, so the arrow is moving and indeed you tell to yourself that this vector field must have a non zero derivative.
Now, same scene, but you turn with the vector, meaning that you always turn your head in order to keep the vector pointing upward to your eyes. Obvisouly, this times there is no movement of the vector, and so the vector field must have a zero derivative.
The first point of view is not possible in an abstract manifold, since there is not a nice way to see all the tangent vectors based on different points (and so living in different tangent spaces) living in a same ambient vector space. So we need to chose something, for example a curve, to keep track of how we are going to connect all these different spaces. This is what leads to your definition of Lie derivative. For example, you can check that the second illustration that I described to you is correct, by showing that $\mathcal{L}_VV=0$ for all vector field $V$.


TITLE: Equivariant homotopy, simplicially
QUESTION [9 upvotes]: It is a classic result of Kan that the homotopy categories (with appropriate model structures) of simplicial sets and of topological spaces (in fact, one could only care about CW-complexes) are equivalent. One could ask if there is a similar result for G-equivariant homotopy category of spaces, where G is a topological group (usually a Lie group) and a morphism of G-spaces is a map, strictly commuting with G-action: $$f(gx) = g\cdot f(x)$$
Since the G-equivariant category admits a very nice theory of cell decompositions and in fact is equivalent to a homotopy category of space presheaves on the category of G-orbits (Elmendorf's theorem), one could guess that such a result should be true. However, I would like to see it not on the level of $\mathrm{SSet}$-presheaves on orbits, but rather directly in the standard definitions. I see 2 problems. Firstly, while cellular and simplicial descriptions of spaces are very similar, they are not equivalent in any sense I could see (besides giving the same homotopy category). The second problem is that if we pass from topological spaces to simplicial sets, we should also change a topological group into a simplicial group. While there is an obvious way to go back and forth (singular complex $\mathrm{Sing}: \mathrm{Top} \to \mathrm{SSet}$ and geometric realization $|\cdot|: \mathrm{SSet} \to \mathrm{Top}$), it is very unobvious that this gives the correct correspondence, since $X \to |\mathrm{Sing}X|$ is just a weak homotopy equivalence, and the equivariant category doesn't respect homotopy equivalences in general, both for spaces and groups. A contractible space can have different non-equivalent G-structures (e.g. free and trivial), and homotopy equivalent (as spaces) groups can give different equivariant categories.
So the question is, can we generalize Kan's theorem to the equivariant setting? If we do, then how does this correspondence look like and how far is it from the intuition? For example, are there finite-dimensional simplicial groups G, such that their G-homotopy categories are not equivalent to equivariant categories of Lie groups? Can we associate with a space $X$ some simplicial group $G_X$, such that the $G_X$-equivariant category is the same as fibrations over $X$?

REPLY [3 votes]: Perhaps you should look at the series of papers by Dwyer and Kan, starting with
W. G. Dwyer and D. M. Kan, Singular functors and realization functors, Nederl. Akad. Wetensch. Indag. Math., 87, (1984), 147 – 153.
I do not think you will find all the answers that you require there, as your reluctance to see things in terms of simplicial presheaves is a hinderance from their point of view, but their discussion (and in later papers by others) may help. In fact I seem to remember that Dwyer and Kan's viewpoint (taken up by Dror later on) implies that there is a variant of the singular complex /  geometric realisation adjointness that does the job. The standard simplicial models are more or less replaced by the $G/H\times \Delta^n$.


TITLE: Regular cuts and complete boolean algebra
QUESTION [0 upvotes]: In Chapter 7 of Jech's Set Theory, he proves that every Boolean algebra has a completion (see this post here for the definition of a Boolean algebra and its partial ordering). The proof is as follows:

Theorem 7.13. Every Boolean algebra has a completion.
Proof. We use a construction similar to the method of Dedekind cuts. Let $A$ be a Boolean algebra. Let us call a set $U\subseteq A^+$ a cut if
$$(7.24) \qquad p\leq q \text{ and } q\in U\text{ implies } p\in U.$$
For every $p\in A^+$, let $U_p$ denote the cut $\{x:x\leq p\}$.
A cut $U$ is regular if
$$\text{whenever } p\not\in U, \text{ then there exists }q\leq p \text{ such that }U_q\cap U=\varnothing.$$
Note that every $U_p$ is regular, and that every cut includes some $U_p$.
We let $B$ be the set of all regular cuts in $A^+$. We claim that $B$, under the partial ordering by inclusion, is a complete Boolean algebra. note that the intersection of any collection of regular cuts is a regular cut, and hence each cut $U$ is included in a least regular cut $\overline{U}$. In fact,
$$\bar U=\{p:(\forall q\leq p) U\cap U_q\neq \varnothing\}.$$
Thus for $u,v\in B$ we have
$$u\cdot v=u\cap v,\qquad u+v=\overline{u+v}.$$
The complement of $u\in B$ is the regular cut
$$-u=\{p:U_p\cap u=\varnothing\}.$$
And, of course, $\varnothing$ and $A^+$ are the zero and the unit of $B$. It is not difficult to verify that $B$ is a complete Boolean algebra, and we leave the verification to the reader. Furthermore, for all $p,q\in A^+$ we have $U_p+U_q=U_{p+q}$, $U_p\cdot U_q=U_{p\cdot q}$, and $-U_p=U_{-p}$. Thus $A$ embeds in $B$ as a dense subalgebra.

I don't think I understand what a regular cut is, or in particular why it is the case that every $U_r$ (where $r\in A^+$; I changed the letter for convenience) is regular.
I'm trying to use Dedekind cuts in the rationals to build intuition, but it seems like no Dedekind cuts are regular, given that any two cuts intersect eventually.
Is there any good intuition behind what it means for a cut to be regular? At this point I'm not really even convinced that regular cuts even exist, much less why the ones highlighted here are regular.

REPLY [2 votes]: One way to start, is to naively try to make completion by cuts "work" and see how the need for a new concept naturally comes up. So let's consider the set $C$ of all cuts of $A^+$ under the $\subseteq$ order and try to find a Boolean completion in there somewhere.
A couple easy facts make us immediately optimistic. First off, a union of cuts is a cut, as is the intersection of cuts, so we are in good shape for meets and joins. And moreover, we have $U_p\cap U_q=U_{p\cdot q}$ and $U_p\cup U_q = U_{p+q},$ so an embedding of $A$ embedding is shaping up nicely too.
Where things stop working so nicely is when we consider complements: the complement of a cut is not necessarily a cut and it's not generally the case that $A^+-U_p = U_{-p}.$ We have a nice structure (in fact, it is a complete Heyting algebra!), but it falls short of being a Boolean algebra. Nonetheless, let's see what we can do to make it work.
We know that if our embedding is to work, we need $-U_p=U_{-p}$. As we remarked previously, using set-theoretic complement $-U_p \ne A^+-U_p$ is wrong... we don't want $-U_p$ to contain literally everything that is not $\le p,$ rather we want it to consist of things that are incompatible with $p$. This is is made more clear by the identity $U_{-p}=\{q\in A^+: q\cdot p = 0 \}$, which just comes from the fact that $q\le - p$ if and only if $q\cdot p = 0.$ The natural generalization of this operation to arbitrary cuts is to let $-U = \{q\in A^+: \forall p\in U\; q\cdot p = 0\},$ or more succinctly, $-U = \{q\in A^+: U_q \cap U = \emptyset\}.$
We want our join operation to give us $U+ - U = A^+,$ but it not necessarily true that $U\cup -U=A^+,$ so the join operation needs to do more than just take the union. The key observation is that while $U_p\cup U_{-p}$ might not contain all of $A^+,$ there is a weaker sense in which it is big: it is dense in $A^+$. This suggests that the correct definition of addition is that $U+V$ is the set of all $p\in A^+$ such that $U\cup V$ is dense below $p$.
Finally, we need addition to satisfy $U+U= U,$ so since $U+U$ is the set of all $p$ that $U$ is dense below, it better be the case that $U$ contains every point it is dense below. That's not true for arbitrary cuts, so we restrict our set of cuts to just the ones for which it is true. These are the regular cuts! (Check that Jech's definition of regular cut is indeed equivalent to the statement that the cut contains every point it is dense below.)
Now we check that it works. Are the $U_p$ regular? Yes, any $q\nleq p$ has an extension incompatible with $p$ (this is just the statement that $A^+$ is a separative partial order), so $U_p$ is not dense below $q$ for any $q\notin U_p.$ Is $U+V$ necessarily regular? Yes, if $U+V$ is dense below $p,$ so is $U\cup V$ so if $p\notin U+V$ then $U+V$ is not dense below $p$. What about if we just let $U\cdot V=U\cap V$? That works fine since if $U\cap V$ is dense below $p$, so are $U$ and $V.$ We can likewise confirm that negation works like we expect, and that Jech's statement that $\overline U$ (the collection of all points $U$ is dense below) is the least regular cut containing $U$ and use that to show that joins (finite or infinite) exist and match up to our definition.

That was more than I planned to write... in case all that the motivation has the opposite of the intended effect, the headline news is:

A cut is regular if it contains every point it is dense below.

A couple closing notes

While our motivation made substantial use of the fact that $A$ is a Boolean algebra, this generalizes to show that any separative partial order densely embeds into a complete Boolean algebra. As remarked above, separativity is what guarantees each $U_p$ is regular.
The construction of the completion is a special case of a more general construction called the regular open algebra. For any topological space $X$, we say that $U\subseteq X$ is regular open if it is equal to the interior of its closure. The collection of all regular open subsets of $X$, ordered by $\subseteq$, is a complete Boolean algebra. Here, the relevant topological space is $P,$ where $U\subseteq P$ is open if and only if $U$ is a cut. The operation $U\mapsto \overline U$ is taking the interior of the closure.


TITLE: Compactness theorem equivalences
QUESTION [2 upvotes]: i have this equivalence to compactness theorem that i have problems to prove:
For every first-order theory $T$, every tuple $x̄$ of distinct variables and all sets $\Phi(x̄),\Psi(x̄)$ of first-order formulas, if $T\vDash\forall\bar{x}(\bigwedge\Phi(x̄)\leftrightarrow\bigvee\Psi(x̄))$ then there are finite sets $\Phi_{0}(x̄)\subseteq\Phi(x̄)$ and $,\Psi_{0}(x̄)\subseteq\Psi(x̄)$ such that $T\vDash\forall\bar{x}(\bigwedge\Phi_{0}(x̄)\leftrightarrow\bigvee\Psi_{0}(x̄))$

REPLY [1 votes]: Write $\neg\Psi$ for the set containing the negation of formulas in $\Psi$. Consider only one implication for the moment. From
(0) $T\vDash\forall x(\bigwedge\Phi(x)\rightarrow\bigvee\Psi(x))$
we obtain
$T \cup \Phi(x) \cup  \neg\Psi(x)\models\bot$
By compactness, for some finite set
$T \cup \Phi_0(x) \cup  \neg\Psi(x)\models\bot$
Therefore
(1) $T \models \forall x [\bigwedge \Phi_0(x) \rightarrow  \bigvee\Psi(x)]$
The converse of (0) is equivalent to $T\vDash\forall x(\bigwedge\neg\Psi(x)\rightarrow\bigvee\neg\Phi(x))$. Therefore the same argument yield
$T \models \forall x [\bigwedge\neg\Psi_0(x) \rightarrow  \bigvee\neg\Phi(x)]$ which is equivalent to
(2) $T \models \forall x [\bigwedge\Phi(x) \rightarrow  \bigvee\Psi_0(x)]$
If you put (0), the converse of (0), (1) and (2) together you obtain
$T \models \forall x [\bigwedge\Phi(x)\rightarrow \bigwedge\Phi_0(x) \rightarrow  \bigvee\Psi(x) \rightarrow\bigwedge\Phi(x) \rightarrow  \bigvee\Psi_0(x)\rightarrow  \bigvee\Psi(x)\rightarrow \bigwedge\Phi(x) ]$
EDIT. Note that we obtain a stronger claim than required. Namely, $\bigwedge\Phi(x)\leftrightarrow \bigwedge\Phi_0(x)$ and $\bigvee\Psi(x)\leftrightarrow \bigvee\Psi_0(x)$.


TITLE: $\langle x | p \rangle = \psi_p(x)$ , What this anything to do with $\psi$ (Notation in R.Shankar)
QUESTION [1 upvotes]: In Principal of Quantum Mechanics R.Shankar Page 137
if we project the eigenvalue equation
$$P|p\rangle = p|p\rangle$$
onto the $X $ basis,we get
$$\langle x|P|p\rangle = p \langle  x |p \rangle$$
or,
$$-i\frac{h}{2\pi}\frac{d\psi_p(x)}{dx}=p\psi_p(x) $$
where, $$\psi_p(x)=\langle x |p \rangle.$$
They use the notation $\psi_p(x)$ for $\langle x |p \rangle$ , I know it is just a notation but $\psi$ has noting to do with  $\langle x |p \rangle$ and $p$ is labeled in $\psi(x)$(=$\langle x|\psi \rangle$ ) which is continuous.
The same is used in page 313 for $\langle x| l_z \rangle $ equating it to $\psi_{l_z}(\rho,\phi)$.

REPLY [1 votes]: A (wave-)function $\psi$ in a Hilbert space, is in fact a vector with infinitely many components. In bra-ket notation we write it as $|\psi \rangle$. Instead the complex number $\psi(x)$ is the component of the vector in the (overcomplete) basis $|x\rangle$ which is the position basis. Such wavefunction represents a particle sitting at position $x$ with certainty. So its components themselves must be
$$
\langle x' | x \rangle = \delta(x'-x).
$$
The number $\psi(x)$, formally, is also the scalar product of $|x\rangle$ with $|\psi\rangle$. Indeed, again formally,
$$
\langle x | \psi\rangle = \int dx' \delta(x'-x) \psi(x') = \psi(x).
$$
In its book Shankar simply uses the (common) notation $|p\rangle$ for the wavefunction $\psi_p$ since he knows it depends on the quantum number $p$.


TITLE: Why are $\lambda$-systems defined with the stronger condition $A\subset B\implies B\setminus A\in\Lambda$
QUESTION [0 upvotes]: A class $\Lambda$ over subsets of $E$ is called a $\lambda$-system if among other conditions the following is fullfilled:
$$(A,B\in\Lambda) \wedge (A\subset B) \implies B\setminus A\in\Lambda$$
My question is if there is a reason why the antecedent of this conditional requires that $A\subset B$.  If we would omit this and define a $\lambda$-system as a A class $\Lambda$ s.t. (among the other two conditions):
$$(A,B\in\Lambda) \implies B\setminus A\in\Lambda$$
Since $(A,B\in\Lambda) \wedge (A\subset B) \implies (A,B\in\Lambda)$, the latter definition would have a weaker antecedent in its conditional. We would also still have that every $\sigma$-algebra is a $\lambda$-system, so what, if there exists any, is the reason to define it like this?

REPLY [1 votes]: If one weakens the condition as said, that is considers sets $\Lambda \subseteq \def\P{\mathfrak P}\P(S)$ that have

(1) $S \in \Lambda$(2) If $A, B \in \Lambda$, then $A \setminus B \in \Lambda$(3) If $(A_n) \in {}^{\mathbf N}\Lambda$ is increasing, then $\bigcup_n A_n \in \Lambda$.

then every $\Lambda$ satisfying (1)-(3) is - of course - a $\lambda$-system, but one has strenghed the conditions, so there may be $\lambda$-systems not satisfying (1)-(3). And there are: Any $\lambda$-system satisfying (2) is a $\pi$-system, due to
$$ A \cap B = A \bigm\backslash (A \setminus B) $$
Hence, any $\Lambda$ satisfying (1)-(3) is in fact a $\sigma$-algebra. Hence (1)-(3) do characterize $\sigma$-algebras. So, if we want to study $\lambda$-systems in their own right, we must stuck with

(1) $S \in \Lambda$(2') If $A, B \in \Lambda$ with $B \subseteq A$, then $A \setminus B \in \Lambda$(3) If $(A_n) \in {}^{\mathbf N}\Lambda$ is increasing, then $\bigcup_n A_n \in \Lambda$.
  or, equivalenty
  (1) $S \in \Lambda$(2'') If $B \in \Lambda$, then $S \setminus B \in \Lambda$(3') If $(A_n) \in {}^{\mathbf N}\Lambda$ is a pairwise disjoint sequence, then $\bigcup_n A_n \in \Lambda$.


TITLE: Critical points of an integrated function
QUESTION [1 upvotes]: Let $$F(x)=-\int_{0}^{x^2}\frac{2}{3+e^t}dt$$
Find all critical points of $F(x)$ and determine whether they are minima, maxima or points of inflection. Prove that $F(300)>F(310)$.
First I differentiated the function and equated to $0$:
$$F'(x) = \frac{-4x}{3+e^{x^2}}=0$$
$\implies x=0$ is a critical point.
Then differentiating twice:
$$F''(x)=\frac{-4(3+e^{x^2})+4x^2(2xe^{x^2})}{(3+e^{x^2})^2}=0$$
But I can't solve this equation (using it for $x=0$ gives that it is a maximum). Is there another way to go? I'm guessing the last part has to do with how the function is decreasing due to some critical points.

REPLY [1 votes]: Note that $F(x)$ is an even function and the integrand is positive over all the real line.


TITLE: If $H\leq G$, then $BH \to BG$ is a fiber bundle with fiber $G/H$
QUESTION [3 upvotes]: Suppose $G$ is a topological group and $H\leq G$ is a closed subgroup. The inclusion $H\to G$ induces a map on classifying spaces $BH\to BG$. I've seen in some sources that $BH\to BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.
I understand the reasoning to be as follows. It's a general fact that $EG\to (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EG\to (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.
It's also a general fact that $(EG)/H \to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.
Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BH\to BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.
How can I reconcile these issues?

REPLY [4 votes]: The statement "$BH \to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.
The homotopy-invariant statement is that the homotopy fiber of the map $BH \to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH \to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.
When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $\bullet \to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $\bullet \to X$ is the loop space $\Omega X$.


TITLE: Is every set a subset of a vector space?
QUESTION [0 upvotes]: I was taught that a functional is a map from a subset (not subspace) of a vector space into the reals, $F: D\subset V \to \mathbb{R}$.
I know there are other definitions, but is there any reason to specify "subset of vector space" in this case? Couldn't one say "set" instead, or are there  sets which you can't include in a vector space?

REPLY [1 votes]: A subset needn't be a subspace. For example, the single point set $\{1\}$ is a subset of $\mathbb R$, but not a subspace of it. The only subspaces of $\mathbb R$ are $\mathbb R$ and $\{0\}$.
One often needs to speak of linearity, so that often gives rise to the requirement that the domain reside within a vector space. Linearity requires things like $$F(a{\bf x} + b{\bf y}) = aF({\bf x}) + bF({\bf y})$$
The expression on the right makes sense, but the expression on the left only  makes sense if there is scalar multiplication and vector addition (otherwise, what would "$a{\bf x} + b{\bf y}$" mean?).


TITLE: Find eigenvalues for $T(f) = \int_{-\infty}^x tf(t)dt$
QUESTION [2 upvotes]: Let $V$ be the linear space of all functions continuous on $(-\infty, \infty)$ and such that that the integral $\int_{-\infty}^x tf(t)\,dt$ exists. If $f \in V$, let $g=T(f)$ be defined as $g(x) = \int_{-\infty}^x tf(t)\,dt$. Prove that every $\lambda < 0$ is an eigenvector and determine the eigenfunctions corresponding to $\lambda$.
We know that $T(f) = \int_{-\infty}^x tf(t)\,dt = \lambda f(x)$. (Not sure if this is right so far.) So know what do I do?

REPLY [1 votes]: You wrote "We know that $T(f) = \int_{-\infty}^x tf(t)\,dt = \lambda f(x)$."
You could say we know that $T(f) = \int_{-\infty}^x tf(t)\,dt$ for every $f$ in the domain, and we know that if $f$ is an eigenfunction with eigenvalue $\lambda$ then the second equality holds.  That doesn't indicate that any such function exists.
Differentiating both sides of the equation
$$
\int_{-\infty}^x tf(t)\,dt=\lambda f(x)
$$
we get
$$
xf(x) = \lambda f'(x)
$$
so that
$$
\frac x \lambda\,dx = \frac{df} f.
$$
Hence
$$
\frac{x^2}{2\lambda} + C = \log |f(x)|.
$$
$$
Ae^{x^2/(2\lambda)} = f(x). \tag 1
$$
This holds if $f$ is an eigenfunction with eigenvalue $\lambda$, so we've shown that if any such eigenfunction exists, then it looks like $(1)$.
Now ask yourself: For which values of $\lambda$ is it the case that $\displaystyle\int_{-\infty}^x t e^{t^2/(2\lambda)} \, dt$ exists for every $x$?


TITLE: Check if we can turn a string into a palindrome by reversing a substring
QUESTION [0 upvotes]: Given a string consisting of lower-case characters from English alphabets, we want to reverse a substring from the string such that the string becomes a palindrome.
Note : A Palindrome is a string which equals its reverse.
We need to tell if some substring exists which could be reversed to convert the string into palindrome.
Example : Let string be "zakdakdz". Then the answer for this string is "yes", since we can reverse the string between indexes 4 and 6 to get zakddkaz
Basic Approach : Try to reverse each and every substring and check if we get palindrome or not. However, this is bad approach for a long string.
So is there a better way to solve it ?

REPLY [2 votes]: Quick heuristic: For very large strings, we can count the frequency of all letters ("a", "b", ...) to quickly weed out strings that cannot possibly be turned into a palindrome.
Let $n$ be the length of the string $S$.
If $n$ is even, then each letter must occur an even number of times. If $n$ is odd, then the same is true except for exactly one letter.
Search for solutions: Let $S_i$ be the i-th letter of $S$ and $S_{a:b}$ the substring from (inclusive) index $a$ to $b$.
Checking a substring $S_{a:b}$:

Is $S_a = S_b$? If yes, check $S_{a+1:b-1}$ instead.

Special case: If this substring has a length of 0 or 1, then $S$ is already a palindrome.

Search for all $S_a$ and $S_b$ in $S_{a+1:b-1}$. If there are none, return false.
Foreach $S_a \in S_{a+1:b-1}$ at index $i$:

Reverse the substring $S_{i:b}$.
Is $S$ now a palindrome? If yes, return true.

Foreach $S_b \in S_{a+1:b-1}$ at index $j$:

Reverse the substring $S_{a:j}$.
Is $S$ now a palindrome? If yes, return true.

Return false.

Now apply the substring algorithm to $S$ as $S_{1:n}$.


TITLE: Computing a conditional expectation
QUESTION [1 upvotes]: I'm trying to compute a conditional expectation. If $(\Omega, \mathcal{F}, P)$ denotes a probability space, and let $A, B\in\mathcal{F}$ with $0<P(B)<1$ and let $\mathcal{G}=\{ B, \Omega\backslash B, \Omega, \emptyset\}$. I am then asked to compute $E(1_A\mid \mathcal{G})$.
I figure this is doable, simply computing the $1_B$- and $1_{\Omega\backslash B}$-parts separately. This would yield
$$E(1_A\mid \mathcal{G})=\frac{P(A\cap B)}{P(B)}1_B+\frac{P(A\cap (\Omega\backslash B))}{P(\Omega\backslash B)}1_{\Omega\backslash B}.$$
Now, this leaves me with two questions: First of all, is this even correct reasoning? And more importantly, if this is correct, how would I go about explaining my reasoning to the professor? Isn't there a more explicit way of computing this expectation?

REPLY [0 votes]: If you could 'guess' that the conditional expectation $E[1_A\mid \mathcal{G}]$ should be of the form (here $B^c=\Omega\setminus B$)
$$
E[1_A\mid \mathcal{G}]=\frac{P(A\cap B)}{P(B)}1_B+\frac{P(A\cap B^c)}{P(B^c)}1_{B^c},
$$
then you could simply check that the right-hand side fulfills the requirements for being the conditional expectation of $1_A$ given $\mathcal{G}$. That is:

Is the right-hand side $\mathcal{G}$-measurable,
Is it $P$-integrable,
Is $E[\text{RHS}\cdot 1_C]=E[1_A\cdot 1_C]$ for all $C\in\mathcal{G}$.

This is pretty straight-forward to check.
If you can't 'guess' that the conditional expectation should be of that form, then you have to start by looking at $\mathcal{G}$. A variable $X$ is measurable with respect to $\mathcal{G}$ if and only if it is constant on $B$ and $B^c$. That is
$$
E[1_A\mid\mathcal{G}]=a\cdot 1_B+b\cdot 1_{B^c}
$$
for some constants $a,b\in\mathbb{R}$. This is also clearly $P$-integrable, so we have to use bullet number 3 above to determine $a$ and $b$. Now, bullet number 3 states that for all $C\in\mathcal{G}$ we have
$$
P(A\cap C)=E[1_A\cdot 1_C]=E[(a\cdot 1_B+b\cdot 1_{B^c})\cdot 1_C]=aP(B\cap C)+bP(B^c\cap C).
$$
Now let $C$ 'walk through' $\mathcal{G}$ and determine $a$ and $b$.


TITLE: Why is it said that light can travel through empty space?
QUESTION [14 upvotes]: Here are a few commonly heard sentences that will make my question clear:
Statements #1
“The Michelson Morley experiment once and for all did away with the ether or the idea that light needs a medium in which to travel”
“Light can travel through empty space”
Statements #2
“Light travels though the electromagnetic field”
“The photon field is only one of the fields for each particle in the standard model.”
“There must be dark matter in the universe for our current model of gravity to make sense.”
“Space is not empty. A quantum foam fills it”
You can see the confusion that arises from the conflicting statements which are commonly heard in physics.
The fact that light travels via an electromagnetic field is in clear contradiction to the statement that light travels through empty space. I suppose we could try to join empty space and the EM field into one thing by saying light is the field. Frankly, that sounds like doublespeak.
What am I missing?
Light is a disturbance in the EM field. There is no empty space. Light travels through a field.
Why do some physicists say light can travel through empty space?

REPLY [5 votes]: I would agree with and repeat the idea that an electromagnetic field is a mathematical construct, not a literal reality, but that doesn't mean it is void of meaning and that you should just "shut the eff up and calculate", either. It is a term in mathematical language that lets us translate and integrate a wide variety of statements such as "this room has a lot of light in it", "there is red light shining here", and more and many more we cannot give straightforwardly in English at all, all into one neat expressive construct. But it definitely refers to real things - you just have to be careful as to what real things, and not to say the "word" itself "exists" out there as a reification. Its relevance to the question is that you can translate the question into this mathematical language and express it there. The reason that doesn't answer it is the same reason asking it in, say, Spanish, doesn't answer it.
But I also say the question is not unanswerable, either. Because there is an answer, and that answer lies in empirical evidence: we say light propagates through empty space because we so far have not found that taking anything out of space that would otherwise be there injures the light or impedes its transmission in any way. Instead, it is the presence of things in space that harms transmission. Though we, of course, cannot at least with out existing technology remove all subtler constituents like dark matter from an area of space, but we have no evidence that to do so would impede light - indeed, were light's propagation to depend on the amount of dark matter present, then it would not be "dark" at all!


text:
\begin{document}

\title{\bf The Norton algebra of a $Q$-polynomial
distance-regular graph
}
\author{
Paul Terwilliger}
\date{}

\maketitle
\begin{abstract} We consider the Norton algebra associated with a $Q$-polynomial primitive idempotent of the adjacency matrix for a distance-regular graph.
We obtain a formula for the Norton algebra product that we find attractive.

\bigskip

\noindent
{\bf Keywords}. Bose-Mesner algebra; Krein parameter; $Q$-polynomial; Leonard system.
\hfil\break
\noindent {\bf 2020 Mathematics Subject Classification}.
Primary: 05C50;
Secondary: 05E30.
 \end{abstract}

\section{Introduction} There is a family of highly regular graphs said to be
distance-regular \cite{banIto, BCN, dkt}.

Examples include
the Johnson graphs \cite[Section~9.1]{BCN},
the Hamming graphs \cite[Section 9.2]{BCN},
the Grassmann graphs \cite[Section~9.3]{BCN},
and the dual polar graphs \cite[Section~9.4]{BCN}. The graphs in these four families are particularly
attractive for several reasons: (i) they have a $Q$-polynomial structure, according to which
their Krein parameters vanish in a certain attractive pattern; (ii) these graphs come with
 a ranked partially ordered set that can be used to analyze the graph.
 \medskip

 \noindent In the analysis of any distance-regular graph $\Gamma$, one often considers
 the eigenspaces for the adjacency matrix $A$ of $\Gamma$.  By \cite{norton}  these eigenspaces
 possess an algebra structure, called the Norton algebra, that is commutative but not necessarily associative.
 The Norton product  $\star$ is described as follows.
Let $X$ denote the vertex set of $\Gamma$.

The  rows and columns of $A$ are indexed by $X$. The matrix $A$ acts on
 a vector space $V$ over $\mathbb R$, consisting of column vectors whose coordinates are indexed by $X$.
 For $x \in X$ let $\hat x$ denote the vector in $V$ that has $x$-coordinate 1 and all other coordinates zero.
 So $\lbrace \hat x | x \in X\rbrace$ is a basis for $V$.  The entry-wise product $\circ: V \times V \to V$ satisfies
 $\hat x \circ \hat y = \delta_{x,y} \hat x$ for all $x,y \in X$. The matrix $A$ is diagonalizable since it is symmetric,
 so $V$ is a direct sum of the eigenspaces of $A$.
 For an eigenspace of $A$, the corresponding
 primitive idempotent $E$  acts as the identity on the eigenspace, and zero on the other eigenspaces of $A$. Thus $E$ is the projection
 from $V$ onto the eigenspace. The  eigenspace is $EV$.
For $u, v \in EV$ we have
 $u \star v = E (u \circ v) $.
 \medskip

 \noindent
  Earlier we mentioned $Q$-polynomial structures.  For a given $Q$-polynomial structure on $\Gamma$
  the adjacency matrix $A$ has
  a distinguished primitive idempotent, said to be $Q$-polynomial.
  Recently, several authors have considered  the Norton algebra $EV$ for a $Q$-polynomial primitive idempotent $E$ of $A$.
  This was done by
 C. Maldonado and D. Penazzi in \cite{Norton},  under the assumption that $\Gamma$ is
 a Johnson graph, Hamming graph, or  Grassmann graph.
 It was done by
 F. Levstein, C. Maldonado, and D. Penazzi in \cite{levNorton},
under the assumption that $\Gamma$ is a dual polar graph.  In both articles the authors compute
$\check u \star  \check v $  for all $u,v \in L_1$, where
$\lbrace {\check u} \vert u \in L_1\rbrace$ is a certain spanning set for $EV$ indexed by the set $L_1$ of rank 1 elements in the associated poset.
The results of \cite{Norton, levNorton} are used by J. Huang
 in  \cite{jia} to investigate the extent to which the Norton product is nonassociative.
 \medskip

\noindent In the present paper we consider the Norton algebra $EV$, where $E$ is a  $Q$-polynomial primitive idempotent of the adjacency matrix $A$ for any distance-regular graph $\Gamma$ with diameter $d\geq 2$.
For all vertices $x,y$ of $\Gamma$ we give an explicit formula for the Norton product $E \hat x \star E\hat y$, in terms of a few eigenvalues $\theta_0, \theta_1, \theta_2$
of $A$
and a sequence of scalars $\lbrace \theta^*_i \rbrace_{i=0}^d$  called the dual eigenvalues of $\Gamma$ associated with $E$.
We give two versions of our formula. The first version is more straightforward. To obtain the second version, we use the balanced set condition
\cite{QPchar, newineq}
to make the symmetry $E\hat x \star E \hat y = E \hat y \star E \hat x$ explicit.
 Our main results are Theorems \ref{thm:main} and  \ref{thm:main2}.
 \medskip

 \noindent The paper is organized as follows.
  In Section 2 we review some basic concepts concerning distance-regular graphs.
 In Section 3 we recall the Norton algebra and obtain the first version of our Norton product formula.
 In Section 4 we use the balanced set condition to obtain the second version of our Norton product formula.
 In Section 5 we remark how certain equations in Sections 3, 4 can be obtained using the theory of Leonard systems.

\section{Preliminaries} In this section we review some basic concepts concerning distance-regular graphs. For more background
information we refer the reader to
\cite{banIto, BCN, dkt}.
\medskip

\noindent
Let $\mathbb R$ denote the real number field.
Let $X$ denote a nonempty finite set.
Let ${\rm Mat}_X(\mathbb R)$ denote the $\mathbb R$-algebra consisting of the matrices that have rows and columns indexed by $X$ and all entries in $\mathbb R$.
Let $I$ denote the identity matrix in  ${\rm Mat}_X(\mathbb R)$,
and let $J$ denote the matrix in  ${\rm Mat}_X(\mathbb R)$ that has all entries 1.
 Let $V=\mathbb R X$ denote the vector space over $\mathbb R$ consisting
of the column vectors that have coordinates indexed by $X$ and all entries in $\mathbb R$. The algebra
${\rm Mat}_X(\mathbb R)$ acts on $V$ by left multiplication.

For $x \in X$  let $\hat x$ denote the vector in $V$ that has $x$-coordinate 1 and all other coordinates 0.
The vectors $\lbrace \hat x | x \in X\rbrace$ form a basis for $V$.
Let $\bf 1$ denote the vector in $V$ that has all entries 1. So ${\bf 1} = \sum_{x \in X} \hat x$.
Note that $J \hat x = \bf 1$ for all $x \in X$.
\medskip

\noindent Let $\Gamma=(X,R)$ denote an undirected connected graph, without loops or multiple edges, with vertex set $X$,
edge set $R$, and path-length distance function $\partial$. Recall the diameter $d={\rm max}\lbrace \partial (x,y) \vert x, y \in X\rbrace$.
For $x \in X$ and an integer $i\geq 0$ define $\Gamma_i(x) = \lbrace y \in X \vert \partial(x,y)=i\rbrace$.  We abbreviate
$\Gamma(x) = \Gamma_1(x)$.
For an integer $k\geq 0$, $\Gamma$ is said to be {\it regular with valency $k$} whenever
$k=\vert \Gamma(x)\vert$ for $x \in X$.
The graph $\Gamma$ is said to be {\it distance-regular} whenever for all integers $h,i,j$ $(0 \leq h,i,j\leq d)$
and all $x,y\in X$ at distance $\partial(x,y)=h$, the scalar

$ p^h_{i,j} = \vert \Gamma_i(x) \cap \Gamma_j(y)\vert$

is independent of $x$ and $y$. The scalars $p^h_{i,j} $ are called the {\it intersection numbers} of $\Gamma$.
For the rest of this paper, we assume that $\Gamma$ is distance-regular with diameter $d\geq 2$.
By construction $p^h_{i,j}=p^h_{j,i}$ for $0 \leq h,i,j\leq d$.
By the triangle inequality we find that for $0 \leq h,i,j\leq d$,
\begin{enumerate}
\item[\rm (i)] $p^h_{i,j}=0$ if one of $h,i,j$ is greater than the sum of the other two;
\item[\rm (ii)] $p^h_{i,j}\not=0$ if one of $h,i,j$ is equal to the sum of the other two.
\end{enumerate}
We abbreviate
$c_i = p^i_{1,i-1}$ $(1 \leq i \leq d)$,
$a_i = p^i_{1,i}$ $(0 \leq i \leq d)$,
$b_i = p^i_{1,i+1}$ $(0 \leq i \leq d-1)$.
By construction $c_i \not=0$ for $1 \leq i \leq d$ and $b_i \not=0$ for $0 \leq i \leq d-1$.
The graph $\Gamma$ is regular with valency $k=b_0$. Moreover
$k = c_i + a_i + b_i$ for $0 \leq i \leq d$,
where $c_0=0$ and $b_d=0$.
\medskip

\noindent Next we recall the Bose-Mesner algebra of $\Gamma$.
For $0 \leq i \leq d$ let $A_i$ denote the matrix in ${\rm Mat}(\mathbb R)$  that has
$(x,y)$-entry
\begin{align*}
(A_i)_{x,y} = \begin{cases}
 1, & {\mbox{\rm if $\partial(x,y)=i$}}; \\
0, &{\mbox{\rm if $\partial(x,y)\not=i$}}
\end{cases}
\qquad \qquad (x,y \in X).
\end{align*}
We call $A_i$ the {\it $i^{\rm th}$ distance-matrix} of $\Gamma$.
Note that $A_0=I$. We abbreviate $A=A_1$ and call this the {\it adjacency matrix} of $\Gamma$.
By the construction,
\begin{align*}
A_i A_j = \sum_{h=0}^d p^h_{i,j} A_h \qquad \qquad (0 \leq i,j\leq d).
\end{align*}
By these comments the matrices $\lbrace A_i \rbrace_{i=0}^d$ form a basis for a commutative subalgebra of ${\rm Mat}_X(\mathbb R)$.
This algebra is denoted by $M$ and called the {\it Bose-Mesner algebra} of $\Gamma$. The algebra $M$ is generated by $A$ \cite[p.~190]{banIto}.
\medskip

\noindent Next we recall the primitive idempotents and eigenvalues of $\Gamma$.
By \cite[p.~45]{BCN} the vector space $M$ has a basis $\lbrace E_i \rbrace_{i=0}^d$ such that (i) $E_0=\vert X \vert^{-1} J$;

(ii) $I = \sum_{i=0}^d E_i$;
(iii) $E_i E_j = \delta_{i,j} E_i $ $ (0 \leq i,j\leq d)$.

This basis is unique up to permutation of $\lbrace E_i \rbrace_{i=1}^d $.
We call $\lbrace E_i \rbrace_{i=0}^d$ the {\it primitive idempotents} of $M$ (or $\Gamma$).
The primitive idempotent $E_0$ is called {\it trivial}.
By construction, there exist real numbers $\lbrace \theta_i\rbrace_{i=0}^d$ such that
$A = \sum_{i=0}^d \theta_i E_i$.
The $\lbrace \theta_i\rbrace_{i=0}^d$ are mutually distinct since $A$ generates $M$.
Using $E_0=\vert X \vert^{-1} J$ we obtain $\theta_0=k$.
The scalars
$\lbrace \theta_i\rbrace_{i=0}^d$ are called the {\it eigenvalues} of $A$ (or $\Gamma$).
\medskip

\noindent Next we recall the Krein parameters of $\Gamma$.
 For $0 \leq i,j\leq d$ we have $A_i \cdot A_j = \delta_{i,j}A_i$, where $\cdot$ denotes
the entry-wise product for ${\rm Mat}_X(\mathbb R)$. Therefore $M$ is closed under the $\cdot$ product. Consequently there exist $q^h_{i,j} \in \mathbb R$ $(0 \leq h,i,j\leq d)$ such that
\begin{align}
\label{eq:eee}
E_i \cdot E_j = \vert X \vert^{-1} \sum_{h=0}^d q^h_{i,j} E_h \qquad \qquad (0 \leq i,j\leq d).
\end{align}
By construction $q^h_{i,j} = q^h_{j,i} $ for $0 \leq h,i,j\leq d$.
By \cite[Proposition~4.1.5]{BCN} we have $q^h_{i,j}\geq 0$ for $0 \leq h,i,j\leq d$. The scalars $q^h_{i,j}$ are called the {\it Krein parameters} of $\Gamma$.
\medskip

\noindent We describe one significance of the Krein parameters. In this description, we will use the following notation.
For $u \in V$ and $x \in X$ let $u_x$ denote the $x$-coordinate of $u$. So $u = \sum_{x \in X} u_x \hat x$.
 For $u, v \in V$ their entry-wise product $u\circ v$ is the vector in $V$ that has $x$-coordinate $u_x v_x$ for all
$x \in X$. So

$u \circ v = \sum_{x \in X} u_x v_x \hat x$.

For $x, y \in X$ we have
\begin{align}
\label{eq:xycirc}
\hat x \circ \hat y = \begin{cases}
 \hat x, & {\mbox{\rm if $x=y$}}; \\
0, &{\mbox{\rm if $x\not=y$.}}
\end{cases}
\end{align}
For $v \in V$ we have ${\bf 1} \circ v = v$.  For subspaces $Y, Z $ of $V$ define
$Y \circ Z = {\rm Span} \lbrace y \circ z \vert y \in Y, z \in Z\rbrace$.
 By \cite[Proposition~5.1]{norton} we have
\begin{align}\label{prop:norton}
E_i V \circ E_j V = \sum_{\stackrel{ \scriptstyle 0 \leq h \leq d }{ \scriptstyle q^h_{ij} \not=0}} E_hV
\qquad \qquad (0 \leq i,j\leq d).
\end{align}

\noindent Next we recall the $Q$-polynomial property.
The given ordering $\lbrace E_i \rbrace_{i=1}^d$ of the nontrivial primitive idempotents of $\Gamma$ is said to be {\it $Q$-polynomial} whenever
for $0 \leq h,i,j\leq d$,
\begin{enumerate}
\item[\rm (i)] $q^h_{i,j}=0$ if one of $h,i,j$ is greater than the sum of the other two;
\item[\rm (ii)] $q^h_{i,j}\not=0$ if one of $h,i,j$ is equal to the sum of the other two.
\end{enumerate}
Let $E$ denote a nontrivial primitive idempotent of $\Gamma$. We say that $E$  is {\it  $Q$-polynomial}
whenever there exists a $Q$-polynomial ordering $\lbrace E_i\rbrace_{i=1}^d$ of the nontrivial primitive idempotents of $\Gamma$
such that $E=E_1$. For the rest of this paper we assume that $E$ is $Q$-polynomial.
By construction, there exist real numbers $\lbrace \theta^*_i\rbrace_{i=0}^d$  such that
\begin{align}
\label{eq:Esum}
E = \vert X \vert^{-1} \sum_{i=0}^d \theta^*_i A_i.
\end{align}
By \cite[p.~260]{banIto} the scalars $\lbrace \theta^*_i \rbrace_{i=0}^d$ are mutually distinct.
The scalars $\lbrace \theta^*_i \rbrace_{i=0}^d$
are called the {\it dual eigenvalues} of $\Gamma$ associated with $E$. For notational convenience let $\theta^*_{-1}$ and $\theta^*_{d+1}$
denote indeterminates.  Taking the trace of each side of (\ref{eq:Esum}) yields $\theta^*_0 = {\rm rank}(E)$. Also,
multiplying both sides of (\ref{eq:Esum}) by $A$ and evaluating the result yields
\begin{align}
c_i \theta^*_{i-1} +
a_i \theta^*_i +
b_i \theta^*_{i+1}
= \theta_1 \theta^*_i
\qquad \qquad (0 \leq i \leq d).
\label{eq:recurse}
\end{align}

\section{The Norton algebra}
\noindent We continue to discuss the distance-regular graph $\Gamma$ and its  $Q$-polynomial primitive idempotent $E$.
In this section we turn the vector space $EV$ into a commutative nonassociative algebra called the Norton algebra.

\begin{definition} {\rm (See \cite[Proposition~5.2]{norton}.)}
\label{def:norton} \rm
The {\it Norton algebra} of $\Gamma$ consists of
the vector space $EV$, together with the product
\begin{align*}

 u \star v = E (u \circ v) \qquad \qquad (u,v \in EV).
 \end{align*}
 \end{definition}
\medskip

\noindent The Norton algebra is commutative, but not necessarily associative.
\medskip

\noindent The vector space $EV$ is spanned by the vectors $\lbrace E\hat x \vert x \in X\rbrace$.
These vectors are nonzero, mutually distinct, and
 linearly dependent \cite[Theorem~1.1]{QPchar}.
As we investigate
the Norton product $\star$ it is natural to consider $E\hat x \star E\hat y$ for all $x,y \in X$.
In the next two lemmas we discuss some extremal cases.

\begin{lemma} \label{lem:warm}
For $x \in X$,
\begin{align}
E \hat x \star E \hat x = |X|^{-1} q^1_{1,1}E \hat x.
\label{eq:ese}
\end{align}
\end{lemma}
\begin{proof} By (\ref{eq:eee}) we have $E \cdot E = \vert X \vert^{-1} \sum_{h=0}^d q^h_{1,1} E_h$.  For this equation, multiply each side by $E$ to obtain
$E (E \cdot E) = \vert X \vert^{-1} q^1_{1,1} E$. For this equation, compare column $x$  of each side to obtain (\ref{eq:ese}).
\end{proof}

\begin{lemma} \label{lem:warmup1}
The following are equivalent:
\begin{enumerate}
\item[\rm (i)] $E\hat x \star E \hat y =0$ for all $x,y\in X$;
\item[\rm (ii)] $u \star v = 0$ for all $u,v \in EV$;
\item[\rm (iii)] The Krein parameter $q^1_{1,1} =0$.
\end{enumerate}
\end{lemma}
\begin{proof} By
(\ref{prop:norton}) and the construction.
\end{proof}

\noindent We have been discussing some extremal cases. Before we proceed to the general case, we bring in some notation.
To motivate things, observe that for $x \in X$ and $0 \leq i \leq d$,
\begin{align}
A_i \hat x = \sum_{z \in \Gamma_i(x)}  {\hat z}.

\label{eq:AiExp}
\end{align}
\begin{lemma}\label{lem:double} For
$x,y \in X$ and $0 \leq i,j\leq d$ we have
\begin{align*}
A_i \hat x \circ A_j \hat y =
\sum_{z \in \Gamma_i(x) \cap \Gamma_j(y)}  {\hat z}.
 \end{align*}
 \end{lemma}
 \begin{proof} Use
 (\ref{eq:xycirc})
  and
 (\ref{eq:AiExp}).
 \end{proof}

\begin{definition}\label{def:xy}
\rm Pick $x, y \in X$ and write $i = \partial (x,y)$. Define
\begin{align}
 x^+_y &= A \hat x \circ A_{i+1} \hat y = \sum_{z \in \Gamma(x) \cap \Gamma_{i+1}(y)}  {\hat z},
 \label{eq:xyp}
\\
x^0_y &= A \hat x \circ A_{i} \hat y =
\sum_{z\in \Gamma(x) \cap \Gamma_i(y)} {\hat z},
\label{eq:xy0}
\\x^-_y &= A \hat x \circ A_{i-1} \hat y =
\sum_{z \in \Gamma(x) \cap \Gamma_{i-1}(y) }  {\hat z},
\label{eq:xym}
 \end{align}
\noindent where we understand
$A_{-1}=0$, $\Gamma_{-1}(x)= \emptyset$ and
$A_{d+1}=0$, $\Gamma_{d+1}(x)=\emptyset$.

\end{definition}
\noindent We clarify the notation (\ref{eq:xyp})--(\ref{eq:xym}).
 Pick $x, y \in X$. If $\partial(x,y)=d$ then $x^+_y = 0$. If $\partial(x,y)=1$ then $x^-_y = \hat y$.  If $x=y$ then $x^0_y=0$ and $x^-_y=0$.

\begin{lemma}
\label{lem:sum} For $x,y \in X$ we have
\begin{align}
x^+_y + x^0_y + x^-_y &= A \hat x,
\label{eq:sum1}
\\
Ex^+_y + Ex^0_y + Ex^-_y &= \theta_1 E \hat x.
\label{eq:sum2}
\end{align}
\end{lemma}
\begin{proof} To verify (\ref{eq:sum1}), note that each side is equal to
$\sum_{z \in \Gamma(x)} \hat z$.
To get (\ref{eq:sum2}), apply $E$ to each side of (\ref{eq:sum1}),  and use $EA = \theta_1 E$.
\end{proof}

 \noindent The following is our first main result.

\begin{theorem}\label{thm:main}
Assume that $\Gamma$ is  $Q$-polynomial with respect to $E$.
Then for all $x,y\in X$ we have
\begin{align}
 E{\hat x}\star E{\hat y} &=
 \frac{ (\theta^*_{i-1}-\theta^*_i) Ex^-_y+(\theta^*_{i+1}-\theta^*_{i})
Ex_y^+ +(\theta_1-\theta_2)\theta^*_i E{\hat x}+(\theta_2-\theta_0)E{\hat y}}
  {\vert X \vert (\theta_1-\theta_2)}
  \label{eq:main}
\end{align}
where $i = \partial (x,y)$. Here $\theta^*_{-1} $ and $\theta^*_{d+1} $ denote indeterminates.
\end{theorem}
\begin{proof} We consider the vector
\begin{align*}
 E (A \hat x \circ E \hat y) -\theta_2
 E (\hat x \circ E \hat y).

 \end{align*}
We evaluate this vector in two ways.
For the first evaluation, use $A-\theta_2 I = \sum_{h=0}^d (\theta_h - \theta_2) E_h$
to obtain
\begin{align*}
 E (A \hat x \circ E \hat y) -\theta_2
 E (\hat x \circ E \hat y)
&=\sum_{h=0}^d (\theta_h-\theta_2) E (E_h \hat x \circ E \hat y).
\end{align*}
For the above sum, we examine the $h$-summand for $0 \leq  h \leq d$.  For $h=0$ the summand is $(\theta_0-\theta_2)\vert X \vert^{-1} E \hat y$ because
\begin{align*}
E_0 \hat x \circ E \hat y = \vert X \vert^{-1} J \hat x \circ E \hat y = \vert X \vert^{-1} {\bf 1} \circ E \hat y = \vert X \vert^{-1} E \hat y.
\end{align*}
For $h=1$ the summand is
$(\theta_1-\theta_2)E\hat x \star E \hat y$ by Definition \ref{def:norton}.
For  $h=2$ the summand is zero by construction. For
$3 \leq h \leq d$ the summand is zero by (\ref{prop:norton}) and the definition of $Q$-polynomial.
 By these comments,
\begin{align}
 E (A \hat x \circ E \hat y) -\theta_2
 E (\hat x \circ E \hat y) =
 (\theta_0-\theta_2) \vert X \vert^{-1} E \hat y + (\theta_1-\theta_2)E \hat x \star E \hat y.
\label{eq:L1}
\end{align}
For the second evaluation, use $E= \vert X \vert^{-1} \sum_{\ell=0}^d \theta^*_\ell A_\ell$
to obtain
\begin{align*}
(A-\theta_2 I) \hat x \circ E \hat y &= \vert X \vert^{-1} \sum_{\ell=0}^d \theta^*_\ell (A-\theta_2 I) \hat x \circ A_\ell \hat y.
\end{align*}
For the above sum, we examine the $\ell$-summand for $0 \leq \ell \leq d$.
The term $A \hat x \circ A_\ell \hat y$ is equal to $x^-_y$ (if $\ell=i-1$) and
$x^0_y$ (if $\ell=i$) and $x^+_y$ (if $\ell=i+1$) and zero
(if $\vert \ell-i\vert > 1$). The term $\hat x \circ A_\ell \hat y$ is equal to
$\hat x$ (if $\ell=i$) and zero (if $\ell\not=i$). By these comments,
\begin{align*}
(A-\theta_2 I) \hat x \circ E \hat y &= \vert X \vert^{-1}(\theta^*_{i-1} x^-_y +
\theta^*_i x^0_y + \theta^*_{i+1} x^+_y - \theta_2 \theta^*_i  \hat x).
\end{align*}
Therefore
\begin{align}
 E (A \hat x \circ E \hat y) - \theta_2 E(\hat x \circ E \hat y)
&= \vert X \vert^{-1}(\theta^*_{i-1} E x^-_y +
\theta^*_i E x^0_y + \theta^*_{i+1} E x^+_y - \theta_2 \theta^*_i E\hat x).
\label{eq:L2}
\end{align}
Comparing (\ref{eq:L1}), (\ref{eq:L2}) we obtain
\begin{align}
\vert X \vert  (\theta_1-\theta_2)E \hat x \star E \hat y
=
\theta^*_{i-1} E x^-_y +
\theta^*_i E x^0_y + \theta^*_{i+1} E x^+_y - \theta_2 \theta^*_i E\hat x
+(\theta_2-\theta_0) E \hat y.
\label{eq:two}
\end{align}
In (\ref{eq:two}), eliminate the term $E x^0_y$  using (\ref{eq:sum2}).

In the resulting equation, solve for $E \hat x \star E \hat y$ and we are done.
\end{proof}

\noindent Referring to Theorem \ref{thm:main},  the formula for $E \hat x \star E \hat y$ can be simplified if $i\in \lbrace 0,1,d\rbrace $.  This simplification is discussed next.
\begin{corollary} \label{cor:01D}
Assume that $\Gamma$ is  $Q$-polynomial with respect to $E$. Then {\rm (i)--(iii)} hold below.
\begin{enumerate}
\item[\rm (i)] For $x \in X$,
\begin{align*}
E \hat x \star E \hat x = \frac{\theta_1 \theta^*_1 - \theta_2 \theta^*_0 + \theta_2 - \theta_0}{\vert X \vert (\theta_1 - \theta_2)} E\hat x.
\end{align*}
\item[\rm (ii)] For $x,y \in X$ at distance $\partial(x,y)=1$,
\begin{align*}
E\hat x \star E \hat y =
\frac{ (\theta^*_2-\theta^*_1) E x^+_y +(\theta_1-\theta_2)\theta^*_1 E \hat x + (\theta_2-\theta_0+\theta^*_0 - \theta^*_1)E \hat y}{\vert X \vert (\theta_1-\theta_2)}.
\end{align*}
\item[\rm (iii)] For $x,y \in X$ at distance $\partial(x,y)=d$,
\begin{align*}
 E{\hat x}\star E{\hat y} &=
 \frac{ (\theta^*_{d-1}-\theta^*_d) Ex^-_y +(\theta_1-\theta_2)\theta^*_d E{\hat x}+(\theta_2-\theta_0)E{\hat y}}
  {\vert X \vert (\theta_1-\theta_2)}.
\end{align*}
\end{enumerate}
\end{corollary}
\begin{proof} (i)  We evaluate (\ref{eq:main}) with $y=x$ and $i=0$. We have $x^-_y=0$ and $x^0_y=0$, so $E x^+_y = \theta_1 E \hat x$ by Lemma \ref{lem:sum}.
\\
\noindent (ii) Set $i=1$ in (\ref{eq:main}) and use $x^-_y = \hat y$.
\\
\noindent (iii) Set $i=d$ in (\ref{eq:main}) and use $x^+_y = 0$.
\end{proof}

 \begin{corollary} \label{cor:q111}
 Assume that $\Gamma$ is $Q$-polynomial with respect to $E$.
 Then the Krein parameter $q^1_{1,1}$  satisfies
 \begin{align*}
 q^1_{1,1} = \frac{ \theta_1 \theta^*_1 - \theta_2 \theta^*_0 + \theta_2 - \theta_0 } {\theta_1-\theta_2}.
 \end{align*}
 \end{corollary}
 \begin{proof} Compare
 Lemma
 \ref{lem:warm} and Corollary  \ref{cor:01D}(i).
 \end{proof}

 \noindent The eigenvalue $\theta_2$ appears in the above results.  By \cite[Lemma~19.21]{LSnotes} we find that
 $1+ \theta_1$, $1+ \theta^*_1$ are nonzero and
 \begin{align}
 \frac{1+ \theta_1}{\theta_0-\theta_2} =
 \frac{1+ \theta^*_1}{\theta^*_0-\theta^*_2}.
\label{eq:th2}

 \end{align}

 \section{The Norton product in symmetric form}

 We continue to discuss  the distance-regular graph $\Gamma$  and its $Q$-polynomial primitive idempotent
$E$. Pick distinct $x, y \in X$. In the formula
 (\ref{eq:main}) we computed
  $E \hat x \star E \hat y$.
We have $E \hat x \star E \hat y=E \hat y \star E \hat x$,
so the right-hand side of   (\ref{eq:main}) must be invariant  if we interchange $x,y$.
In this section, we express the right-hand side of (\ref{eq:main}) in a form that makes this
invariance explicit. We will use a result known as the balanced set condition.

\begin{lemma} \label{lem:bal} {\rm (See \cite[Theorem~1.1]{QPchar}, \cite[Theorem~3.3]{newineq}.)} For distinct $x,y \in X$ we have
\begin{align}
Ex^-_y - E y^-_x &= c_i \frac{\theta^*_1-\theta^*_{i-1}}{\theta^*_0-\theta^*_i} (E \hat x - E \hat y),
\label{eq:bsc1}
\\
 Ex^+_y - E y^+_x &= b_i \frac{\theta^*_1-\theta^*_{i+1}}{\theta^*_0-\theta^*_i} (E \hat x - E \hat y),
 \label{eq:bsc2}
 \end{align}
 where $i=\partial(x,y)$.
 \end{lemma}

 \begin{corollary}
 \label{cor:CB}  For distinct $x,y \in X$ we have
 \begin{align*}
 C(x,y) = C(y,x), \qquad \qquad B(x,y)= B(y,x)
 \end{align*}
 \noindent where
 \begin{align}
 C(x,y) &=
Ex^-_y -  c_i \frac{\theta^*_1-\theta^*_{i-1}}{\theta^*_0-\theta^*_i} E \hat x,
\label{eq:CC1}
\\
B(x,y) &= Ex^+_y -  b_i \frac{\theta^*_1-\theta^*_{i+1}}{\theta^*_0-\theta^*_i} E \hat x
\label{eq:BB1}
 \end{align}
 and $i = \partial(x,y)$.
\end{corollary}
\begin{proof} Rearrange the terms in (\ref{eq:bsc1}), (\ref{eq:bsc2}).
\end{proof}
\noindent We clarify the meaning of (\ref{eq:CC1}) and (\ref{eq:BB1}).
For $i=1$ we have $C(x,y) = E\hat x + E \hat y$. For $i=d$ we have $B(x,y)=0$.
\medskip

\noindent For distinct $x,y \in X$ we are going to express $E\hat x \star E \hat y$ in terms of $C(x,y)$ and $B(x,y)$.
The following equation will be useful.

 \begin{lemma} \label{cor:cibi}
 We have
 \begin{align*}
 c_i  \frac{ (\theta^*_1 - \theta^*_{i-1})(\theta^*_{i-1} - \theta^*_i)}{\theta^*_0-\theta^*_i}
 + b_i  \frac{ (\theta^*_1 - \theta^*_{i+1})(\theta^*_{i+1} - \theta^*_i)}{\theta^*_0-\theta^*_i}
= (\theta_2 - \theta_1) \theta^*_i + \theta_2 - \theta_0
\end{align*}
for $1 \leq i \leq d-1$ and
 \begin{align*}
 c_d \frac{ (\theta^*_1 - \theta^*_{d-1})(\theta^*_{d-1} - \theta^*_d)}{\theta^*_0-\theta^*_d}
= (\theta_2 - \theta_1) \theta^*_d + \theta_2 - \theta_0.
\end{align*}
\end{lemma}
\begin{proof}  In the equation $0 = E\hat x \star E \hat y - E\hat y \star E \hat x$, expand the right-hand side using
Theorem \ref{thm:main} and evaluate the result using
Lemma \ref{lem:bal}. Examine the outcome using the fact that $E \hat x \not=E \hat y$.
\end{proof}
\noindent The following is our second main result.
\begin{theorem}\label{thm:main2}
Assume that $\Gamma$ is $Q$-polynomial with respect to $E$. Then for distinct $x,y \in X$ we have
\begin{align}
 E{\hat x}\star E{\hat y} &=
 \frac{ (\theta^*_{i-1}-\theta^*_i) C(x,y)+(\theta^*_{i+1}-\theta^*_{i})
B(x,y) +(\theta_2-\theta_0)(E \hat x + E{\hat y})}
  {\vert X \vert (\theta_1-\theta_2)}
  \label{eq:main2}
\end{align}
where $i = \partial (x,y)$. Here  $\theta^*_{d+1} $ denotes an indeterminate.
\end{theorem}
\begin{proof} To verify (\ref{eq:main2}), expand the right-hand side using
(\ref{eq:CC1}), (\ref{eq:BB1}) and evaluate the result using
Theorem \ref{thm:main} along with
Lemma \ref{cor:cibi}.
\end{proof}
\noindent Referring to Theorem \ref{thm:main2},  the formula for $E \hat x \star E \hat y$ can be simplified if $i\in \lbrace 1,d\rbrace$.  This simplification is discussed next.

\begin{corollary} \label{cor:1D}
Assume that $\Gamma$ is  $Q$-polynomial with respect to $E$. Then {\rm (i), (ii)} hold below.
\begin{enumerate}
\item[\rm (i)] For  $x,y \in X$ at distance $\partial(x,y)=1$,
\begin{align*}
E\hat x \star E \hat y =
\frac{ (\theta^*_2-\theta^*_1) B(x,y) + (\theta_2-\theta_0+\theta^*_0 - \theta^*_1)(E\hat x + E \hat y)}{\vert X \vert (\theta_1-\theta_2)}.
\end{align*}
\item[\rm (ii)] For $x,y \in X$ at distance $\partial(x,y)=d$,
\begin{align*}
 E{\hat x}\star E{\hat y} &=
 \frac{ (\theta^*_{d-1}-\theta^*_d) C(x,y) +(\theta_2-\theta_0)(E\hat x + E{\hat y})}
  {\vert X \vert (\theta_1-\theta_2)}.
\end{align*}
\end{enumerate}
\end{corollary}
\begin{proof}  (i) Set $i=1$ in (\ref{eq:main2}) and use $C(x,y)= E \hat x + E \hat y$.
\\
\noindent (iii) Set $i=d$ in (\ref{eq:main2}) and use $B(x,y)=0$.
\end{proof}

\section{Remarks}

\noindent
 The algebraic structure of a $Q$-polynomial distance-regular graph can be described using the concept of a Leonard system \cite[Definition~1.4]{ter2};
 this concept was motivated by a theorem of D.~A.~Leonard  \cite[p.~260]{banIto}, \cite{leonard}.
We refer the reader to \cite{ter2, LSnotes} for the standard notation and basic results about Leonard systems.
The equations below are routinely obtained using the formulas in \cite[Sections~19,~20]{LSnotes}.
Let $\Phi$ denote  any Leonard system with diameter $d\geq 2$. For $\Phi$ we have
\begin{align*}
a^*_1 = \frac{(\theta_1-a_0)\theta^*_1-(\theta_2-a_0)\theta^*_0+(\theta_2-\theta_0) c^*_1}{\theta_1-\theta_2}.
\end{align*}
\noindent If we set $a^*_1=q^1_{1,1}$ and $a_0=0$ and $c^*_1=1$  then we recover the formula in
Corollary
\ref{cor:q111}.
\medskip

\noindent For $\Phi$ we also have
\begin{align*}
\frac{c_1 - a_0 + \theta_1}{\theta_0-\theta_2} =
\frac{c^*_1 - a^*_0 + \theta^*_1}{\theta^*_0-\theta^*_2}.
\end{align*}
\noindent If we set $c_1=1$, $a_0=0$ and $c^*_1=1$, $a^*_0=0$ then we recover
(\ref{eq:th2}).
\medskip

\noindent For $\Phi$ we also have
\begin{align*}
 c_i  \frac{ (\theta^*_1 - \theta^*_{i-1})(\theta^*_{i-1} - \theta^*_i)}{\theta^*_0-\theta^*_i}
 + b_i  \frac{ (\theta^*_1 - \theta^*_{i+1})(\theta^*_{i+1} - \theta^*_i)}{\theta^*_0-\theta^*_i}
= (\theta_2 - \theta_1) (\theta^*_i -a^*_0)+ (\theta_2 - \theta_0)c^*_1
\end{align*}
for $1 \leq i \leq d-1$ and
 \begin{align*}
 c_d \frac{ (\theta^*_1 - \theta^*_{d-1})(\theta^*_{d-1} - \theta^*_d)}{\theta^*_0-\theta^*_d}
= (\theta_2 - \theta_1) (\theta^*_d-a^*_0) +(\theta_2 - \theta_0)c^*_1.
\end{align*}
\noindent If we set $a^*_0=0$ and $c^*_1=1$ then we recover the formulas in Lemma \ref{cor:cibi}.
\medskip

\section{Acknowledgement} The author thanks Jia Huang and Kazumasa Nomura for giving this paper a close
reading and offering valuable comments.

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