TITLE: If $\operatorname{Hom}(\delta_V, \delta_W) = 0$, then $\mathfrak{C}(\delta_V) \cap \mathfrak{C}(\delta_W) = 0.$
QUESTION [4 upvotes]: Let $(A, \Delta)$ be a Hopf $^*$-algebra and $\delta_V: V \to V \otimes A$ and $\delta_W: W \to W \otimes A$ be two corepresentations of $(A, \Delta).$
Assume that the space of intertwiners $\operatorname{Hom}(\delta_V, \delta_W) = 0$, then is it true that $\mathfrak{C}(\delta_V) \cap \mathfrak{C}(\delta_W) = 0$, where
$$\mathfrak{C}(\delta) = \operatorname{span}\{(f \otimes\operatorname{id}_A)(\delta(v)): f \in V', v \in V\}$$
denotes the space of matrix coefficients of the corepresentation $(V, \delta).$

REPLY [2 votes]: As mentioned, the present question is related to Question on characterising CQGs. Since the questions are related, the answers are also.
I will work in the setting of Theorem 3.2.12 from Timmerman's book "An invitation to quantum groups and duality".

Let $(A,\Delta)$ be the given Hopf $*$-algebra.
As in (v) in Theorem 3.2.12, let $(\delta_\alpha)_{\alpha\in I}$ us fix an exhaustive set of irreducible unitary corepresentations
$\delta_\alpha:U_\alpha\to U_\alpha\otimes A$, which are pairwise inequivalent. They are indexed by some set $I$.
(As in Proposition 3.2.3 in loc. cit. we have a suitable inner product on $U_\alpha$ making $\delta_\alpha$ unitary.)
Let us use alternative notations
$C(V)=\mathfrak C(\delta_V)$,
$C(W)=\mathfrak C(\delta_W)$.
Since
$\Delta C(V)\subseteq C(V)\otimes C(V)$,
$\Delta C(W)\subseteq C(W)\otimes C(W)$,
we obtain restricted corepresentations of $\Delta:A\to A\otimes A$:
$$
\begin{aligned}
\Delta\Big|_{C(V)} &: C(V)\to C(V)\otimes A\ ,
\\
\Delta\Big|_{C(W)} &: C(W)\to C(W)\otimes A\ ,
\ .
\end{aligned}
$$
Let now $V_1\ne 0$ be a non-trivial minimal invariant subspace of $C(V)$.
So $\Delta\Big|_{V_1}:V_1\to V_1\otimes A$ is irreducible, thus isomorphic to some $\delta_{\alpha_1}$ for some $\alpha_1=\alpha(V_1)$ in the index list $I$.
Proposition 3.2.11 (mentioned in the other answer) shows then
$$
\Delta\Big|_{V_1}\cong\delta_{\alpha_1}^{\boxplus n(\alpha_1)}\ .
$$
Now i need finite dimensionality of $V$, or some Hermitian product (so that orthocomplements of invariant subspaces can be built, and are again invariant) in order to write $V$ as a direct sum of spaces $V_1,V_2,\dots$ - and in this case we have for instance
$$
\begin{aligned}
&\mathfrak C(\delta|_{V_1\oplus V_2})
\\
&=
\operatorname{Span}\{\
(f \otimes\operatorname{id}_A)\delta(v)\ :\
f \in (V_1\oplus V_2)'\ ,\ v \in V_1\oplus V_2\
\}
\\
&=
\operatorname{Span}\{\
((f_1\oplus f_2) \otimes\operatorname{id}_A)\delta(v_1\oplus v_2)\ :\
f_1\in V_1'\ , f_2\in V_2'\ ;\ v_1 \in V_1\ ,\ v_2\in V_2\
\}
\\
&=
\operatorname{Span}\{\
(f_1\otimes\operatorname{id}_A)\delta(v_1)
+
(f_2\otimes\operatorname{id}_A)\delta(v_2)
\ :\
f_1\in V_1'\ , f_2\in V_2'\ ;\ v_1 \in V_1\ ,\ v_2\in V_2\
\}
\\
&=\mathfrak C(\delta|_{V_1}) + \mathfrak C(\delta|_{V_2})
\ .
\end{aligned}
$$
Some (transfinite) inductive argument should give then
$$
C(V)
=\mathfrak C(\delta)
=\mathfrak C(\delta|_V)
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta|_{V_1})
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha_1}^{\boxplus n(\alpha_1)})
=\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha_1})\ .
$$
(Equivalent corepresentations have the same matrix elements.)
A similar relation holds for $W$, so:
$$
\begin{aligned}
\mathfrak C(\delta_V)=
C(V) &=
\sum_{\substack{V_1\subseteq V\\V_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha(V_1)})
=\sum_{\alpha\in I(V)}
\mathfrak C(\delta_{\alpha})
=\bigoplus_{\alpha\in I(V)}
\mathfrak C(\delta_{\alpha})
\subseteq
\bigoplus_{\gamma\in I}
\mathfrak C(\delta_{\gamma})
\ ,
\\
\mathfrak C(\delta_W)=
C(W) &= \sum_{\substack{W_1\subseteq W\\W_1\ne 0\text{ minimal}}}
\mathfrak C(\delta_{\alpha(W_1)})
=\sum_{\beta\in I(W)}
\mathfrak C(\delta_{\beta})
=\bigoplus_{\beta\in I(W)}
\mathfrak C(\delta_{\beta})
\subseteq
\bigoplus_{\gamma\in I}
\mathfrak C(\delta_{\gamma})
\ ,
\end{aligned}
$$
where $I(V)$, $I(W)$ are subsets of the total index set $I$.
If there is now an element $u$ in the intersection of the above two spaces,
then there is a common $\gamma\in I(V)\cap I(W)$, we associate the corresponding minimal spaces $V_1\subseteq C(V)\subseteq A$, $W_1\subseteq C(W)\subseteq A$, and obtain a non-trivial intertwiner after writing
$\Delta|_{V_1}\cong\delta_\gamma^{\boxplus n(V_1)}$,
$\Delta|_{W_1}\cong\delta_\gamma^{\boxplus n(W_1)}$.
The projectors on the $\boxplus$-pieces are also intertwiners, so using kernels and orthocomplements we find a "same piece" $\delta_\gamma$ inside both $\Delta|_{V_1}$ and $\Delta|_{W_1}$.
Despite of the chosen notations, $V_1$ lives inside $C(V)\subseteq A$, not inside $V$. To connect with $V$, we use for some suitable $f\ne 0$ the map $T_f$ between corepresentations:
$\require{AMScd}$
\begin{CD}
@. V @>\delta_{V}>> V\otimes A\\
@. @V\delta_{V} VV @VV\delta_{V}\otimes \operatorname{id}_AV\\
@. V\otimes A @>\operatorname{id}_V\otimes\Delta>> V\otimes A\otimes A\\
@. @Vf\otimes \operatorname{id}_A VV @VV f\otimes \operatorname{id}_A\otimes \operatorname{id}_AV\\
@. \Bbb C\otimes A @>1\otimes\Delta>> \Bbb C\otimes A\otimes A\\
@. @| @|\\
u\in V_1\subseteq C(V)
@>\subseteq>> A @>\Delta>> A\otimes A\\
\end{CD}
Here, $T_f$ is the composition of the maps in the left vertical column. (It is mentioned in loc. cit. at the end of §3.1.1.) And $f$ is chosen so that $u$ is in the image of $T_f$. Using $T_f$ as a bridge, building kernel
and orhtocomplement, we manage get "a piece" $\delta_\gamma$ inside $V$ (i.e. inside $\delta_V$).
A similar argument realizes "a piece" $\delta_\gamma$ inside $W$.
Projectors onto $\boxplus$-pieces can be used to construct intertwiners, so we obtain an intertwiner between $\delta_V$ and $\delta_W$ using the identity of $\delta_\gamma$ realized as above "as a piece" inside them.
This contradicts the assumption $\operatorname{Hom}(\delta_V, \delta_W) = 0$.

As mentioned in the other answer, i am far away from home. The above argument is somehow K-theoretical, splits as much as possible the linear algebra into the corepresentative atoms $(\delta_\alpha)_{\alpha\in I}$, and builds bridges to them. I saw no direct path (from $V$ to $W$) for a solution. Hope that the arguments are OK.


TITLE: Why is $X^2+1$ reducible in $F_2$?
QUESTION [2 upvotes]: One of my modules contains parts of ring theory which I have not done in over a year. So I appreciate this question may seem very basic as I am giving myself a crash course in ring theory by studying examples.
Why is $X^2+1$ reducible in $F_2$?
And why is $X^2-X+1$ irreducible in $F_2$?

REPLY [2 votes]: $X^2 + 1$ is reducible because it is the product of smaller degree polynomials: $(x+1)(x+1) = x^2 + 1$.
Now, if $X^2 -X  + 1 $ was reducible, then it would be the product of two linear factors. That means that the polynomial would have a root in $F_2$. But, you can check (easily) that the polynomial does not have any roots.


TITLE: Proving the given inequality.
QUESTION [1 upvotes]: $z$ is any complex number , we need to prove that :
$|z^{4} + 5z^{2} +4|\geq ||z^{4}|-5|z^{2}|+4|$.
What I did :
I made use of the reverse triangle inequality : $|x-y|\geq ||x|-|y||$.
$|z^{4} + 5z^{2} +4| = |z^{4} + 5z^{2} -(-4)| \geq ||z^{4} + 5z^{2}| -4||$.
Now , $|z^{4} + 5z^{2}|=|z^{4} -(- 5z^{2})| \geq ||z^{4}| - |5z^{2}||$ , Combining this with the previous step , I got :
$||z^{4} + 5z^{2}| -4|| \geq ||z^{4}| - |5z^{2}|-4| = ||z^{4}| - 5|z^{2}|-4|$.
I am getting a negative sign before $4$ , could anyone tell what am I doing wrong ?

REPLY [0 votes]: Hint:
$$|z^{4} + 5z^{2} +4| = |z^{4} + 4 - (-5z^{2})|$$


TITLE: Feynman loop diagrams physical meaning
QUESTION [6 upvotes]: I'm relatively new to Feynman diagrams and have a question about loops. Consider the following diagram:

I understand the electron positron annihilation resulting in a photon but I do not understand the loop in the middle. I've read that the integral is convergent for the loop but I'm looking for a physical explanation. What physical phenomena is happening at the loop part?

REPLY [9 votes]: I am assuming that you are familiar with quantum mechanics. If not, ignore this answer.
Feynman diagrams are a way to keep track of the various terms in the perturbative expansion. Perturbation theory is not something of quantum field theory only, it belongs to many areas of physics, especially quantum mechanics.
The quantum mechanical equivalent would be time-dependent perturbation theory. The transition amplitude from an unperturbed state  $|\alpha\rangle$ at time $t_0$ to a state $|\beta\rangle$ at time $t$ is given by
$$
\mathcal{A}_{\alpha\to\beta} = \langle \alpha|U|\beta\rangle\,,
$$
where $U$ is given by (assuming $\langle\alpha|\beta\rangle=0$ at $t=t_0$ for simplicity)
$$
U \simeq \lambda \frac{\langle \alpha | V | \beta\rangle}{E_\alpha-E_\beta} + \lambda^2 \color{purple}{\sum_n} \frac{\langle \alpha | V |\color{purple}n\rangle\langle \color{purple}n | V | \beta\rangle}{(E_\alpha-E_{\color{purple}n})(E_{\color{purple}n}-E_\beta)} + \cdots,
$$
where $\lambda V$ is the perturbation. (I am being schematic here just to get the point across.)
Now you might say: ok, but where are the loops here?
The ''loop'' is the sum over $\boldsymbol{n}$.
Perturbation theory tells us that in order to compute the contribution of the perturbation $\lambda V$ to the transition amplitude, we should take into account all the states of the theory. Higher order corrections in $\lambda$ result in more and more sums over the states $|n\rangle$.
A loop in a Feynman diagram represents exactly that. We are summing over all the intermediate states which give a non-zero matrix element with $V$ and the external states.
More precisely:

What are the states we sum over? All particle or multi-particle states with any possible momentum. This means that $\sum_n$ will actually be promoted to a sum and an integral over the momentum $\mathrm{d}^4k$. (This integral is the source of all troubles)

How do we see when $\langle\alpha|V|n\rangle$ is non-zero? This information is encoded in the vertices of the diagrams. For instance, the vertex of QED tells you that
$$
\langle \mathrm{photon} | V |\mathrm{positron}+\mathrm{electron}\rangle \neq 0
$$
because you have a vertex with a wiggly leg (photon) and two opposite arrow legs ($e^+$ and $e^-$).

What is the "unperturbed" theory and the perturbation here? The unperturbed theory is the theory of free (non-interacting) photons positrons and electrons, while the perturbation is the above-mentioned potential.

Note: I got the feeling that you were looking for a more ''dynamical'' interpretation, like the photon decays in two electrons which then annihilate into a photon again. Well, this kind of interpretation is unhelpful at best and downright misleading at worst. None of these processes actually take place in any measurable way (photons can't decay!) so it's kind of pointless to pretend that they happen.


TITLE: Discrete logarithm tables for the fields $\Bbb{F}_8$ and $\Bbb{F}_{16}$.
QUESTION [29 upvotes]: The smallest non-trivial finite field of characteristic two is
$$
\Bbb{F}_4=\{0,1,\beta,\beta+1=\beta^2\},
$$
where $\beta$ and $\beta+1$ are primitive cubic roots of unity, and zeros of the
polynomial $x^2+x+1$. Here the multiplication table is given once we know
how to write the non-zero elements as powers of $\beta$. Extend the idea to fields of eight and sixteen elements.
Those fields can be constructed as
$$
\Bbb{F}_8=\Bbb{F}_2[\alpha], \quad\text{and}\quad \Bbb{F}_{16}=\Bbb{F}_2[\gamma],
$$
where $\alpha$ has minimal polynomial $x^3+x+1$, and $\gamma$ has minimal polynomial $x^4+x+1$, both irreducible in $\Bbb{F}_2[x]$.
Task:

Calculate the tables for base $\alpha$ discrete logarithm of $\Bbb{F}_8$ and base $\gamma$ discrete logarithm of $\Bbb{F}_{16}$.

REPLY [30 votes]: A (base-$g$) discrete logarithm of a finite field $\Bbb{F}_q$, is a function
$$
\log_g:\Bbb{F}_q^*\to\Bbb{Z}_{q-1}
$$
defined via the equivalence $g^j=x\Leftrightarrow \log_g(x)=j$. For this to be well-defined it is imperative that $g$ is a primitive element, i.e. a generator of $\Bbb{F}_q^*$, and that the domain of $\log_g$ is the
residue class ring of integer modulo $q-1$, as $g^{q-1}=g^0=1$.
It immediately follows that the discrete logarithm satisfies the familiar rules
$$
\begin{aligned}
\log_g(x\cdot y)&=\log_g(x)+\log_g(y),\\
\log_g(x^n)&=n\cdot\log_g(x)
\end{aligned}
$$
for all elements $x,y\in \Bbb{F}_q^*$ and all integers $n$. The arithmetic
on the r.h.s. is that of the ring $\Bbb{Z}_{q-1}$.

It is known that when $q=8$, a zero $\alpha$ of $x^3+x+1$ generates $\Bbb{F}_8^*$. This is proven by the following calculation, where we repeatedly
use the fact that we are working in characteristic two, and that we have the
relation $\alpha^3=\alpha+1$.
$$
\eqalign{
\alpha^0&=&&=1,\\
\alpha^1&=&&=\alpha,\\
\alpha^2&=&&=\alpha^2,\\
\alpha^3&=&&=1+\alpha,\\
\alpha^4&=&\alpha\cdot\alpha^3=\alpha(1+\alpha)&=\alpha+\alpha^2,\\
\alpha^5&=&\alpha\cdot\alpha^4=\alpha(\alpha+\alpha^2)=\alpha^2+\alpha^3=\alpha^2+(1+\alpha)&=1+\alpha+\alpha^2,\\
\alpha^6&=&\alpha\cdot\alpha^5=\alpha(1+\alpha+\alpha^2)=\alpha+\alpha^2+\alpha^3=
\alpha+\alpha^2+(1+\alpha)&=1+\alpha^2,\\
\alpha^7&=&\alpha\cdot\alpha^6=\alpha(1+\alpha^2)=\alpha+\alpha^3=\alpha+(1+\alpha)&=1.
}$$
We see from the end results in the last column that all the non-zero
quadratic polynomials evaluated at $\alpha$ appear. This is yet another confirmation of the fact that $\alpha$ is a primitive element.
The discrete logarithm is used to replace the cumbersome multiplication (and raising
to an integer power) of the field with more familiar integer arithmetic. Exactly like the old-timers used logarithm tables to replace the error-prone multiplication with the easier addition.
For example
$$
(1+\alpha)(1+\alpha+\alpha^2)=\alpha^3\cdot\alpha^5=\alpha^8=\alpha^7\cdot\alpha=\alpha.
$$
Note that both the base-$\alpha$ discrete logarithms and its inverse mapping are needed. I generate such a table as a part of the program initialization, whenever I carry out extensive computer-aided calculations involving a finite field. The above table gives the discrete logarithm when read from right to left, and the inverse mapping (that we actually produced above) when read from left to right.

Similarly with $q=16$ we use $\gamma$, a zero of $x^4+x+1$. This time the table looks like
$$
\begin{aligned}
\gamma^0&=&1\\
\gamma^1&=&\gamma\\
\gamma^2&=&\gamma^2\\
\gamma^3&=&\gamma^3\\
\gamma^4&=&\gamma+1\\
\gamma^5&=\gamma(\gamma+1)=&\gamma^2+\gamma\\
\gamma^6&=\gamma(\gamma^2+\gamma)=&\gamma^3+\gamma^2\\
\gamma^7&=\gamma^4+\gamma^3=&\gamma^3+\gamma+1\\
\gamma^8&=(\gamma^4)^2=&\gamma^2+1\\
\gamma^9&=\gamma(\gamma^2+1)=&\gamma^3+\gamma\\
\gamma^{10}&=\gamma^4+\gamma^2=&\gamma^2+\gamma+1\\
\gamma^{11}&=&\gamma^3+\gamma^2+\gamma\\
\gamma^{12}&=\gamma^4+\gamma^3+\gamma^2=&\gamma^3+\gamma^2+\gamma+1\\
\gamma^{13}&=\gamma^4+\gamma^3+\gamma^2+\gamma=&\gamma^3+\gamma^2+1\\
\gamma^{14}&=\gamma^4+\gamma^3+\gamma=&\gamma^3+1\\
(\gamma^{15}&=\gamma^4+\gamma=&1)
\end{aligned}
$$
Thus for example
$$
(\gamma^3+1)(\gamma^2+1)=\gamma^{14}\cdot\gamma^8=\gamma^{22}=\gamma^7=\gamma^3+\gamma+1.
$$

As another example of the use of this table I want to discuss the problem of factorizing $x^4+x+1$ over $\Bbb{F}_4$. To that end we need to first identify a copy of $\Bbb{F}_4$ as a subfield of $\Bbb{F}_{16}$. We just saw that $\gamma$ is of order fifteen. Therefore $\gamma^5=\gamma^2+\gamma$ and $\gamma^{10}=\gamma^2+\gamma+1$ are third roots of unity. It is then trivial to check that we have a homomorphism of fields $\sigma:\Bbb{F}_4\to\Bbb{F}_{16}$ given by $\sigma(\beta)=\gamma^5$. Note that composing this (from either end) by the Frobenius automorphism gives an alternative embedding $\beta\mapsto \gamma^{10}$.
Basic Galois theory tells us that
$$
x^4+x+1=(x-\gamma)(x-\gamma^2)(x-\gamma^4)(x-\gamma^8)
$$
as we get the other roots by repeatedly applying the Frobenius automorphism $F:x\mapsto x^2$. Here we see that the factor
$$
(x-\gamma)(x-\gamma^4)=x^2+x(\gamma+\gamma^4)+\gamma^5=x^2+x+\gamma^5
$$
is stable under the automorphism $F^2$, and thus (as we also see directly!) has its
coefficients in the subfield $\sigma(\Bbb{F}_4)$. The same holds for the remaining factor
$$
(x-\gamma^2)(x-\gamma^8)=x^2+x(\gamma^2+\gamma^8)+\gamma^{10}=x^2+x+\gamma^{10}.
$$
Pulling back the effect of $\sigma$ we get the desired factorization
$$
x^4+x+1=(x^2+x+\beta)(x^2+x+\beta+1)
$$
in $\Bbb{F}_4[x]$.

Here is a local version of similar tables for $\Bbb{F}_{256}$


TITLE: Evaluate $\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}$
QUESTION [3 upvotes]: Problem: evaluate $\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}$.
My work: since $n+1 \le k \le 2n$, it is $\frac{1}{2n} \le \frac{1}{k} \le \frac{1}{n+1}$.
Since $k \ge 1$, the exponential with base $k$ is increasing and so, from $\frac{1}{2n} \le \frac{1}{k} \le \frac{1}{n+1}$, it follows that $k^{1/2n} \le k^{1/k} \le k^{1/(n+1)}$.
Since the $\alpha$-th ($\alpha$ real) power is increasing when its base is nonnegative and $k \ge 1$, it follows that $(n+1)^{1/2n} \le k^{1/2n}$ and $k^{1/(n+1)} \le (2n)^{1/(n+1)}$; in conclusion, it is $(n+1)^{1/2n} \le k^{1/k}\le(2n)^{1/(n+1)}$ for any $n+1 \le k \le 2n$ and for any $n\in\mathbb{N}$.
Hence
$$\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k} \ge \lim_{n \to \infty} \prod_{k=n+1}^{2n} (n+1)^{1/2n}=\lim_{n\to\infty}\sqrt{n+1}=\infty$$
$$\implies \lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}=\infty$$
Is this correct? I'm not sure that what I've written about the estimation I get using the fact that the exponential and the real power are increasing.

REPLY [3 votes]: This is right, yes. If you're worried about its rigour, you can take limits as late as possible, viz.$$k^{1/k}\ge k^{1/(2n)}\ge(n+1)^{1/(2n)}\implies\prod_{k=n+1}^{2n}k^{1/k}\ge\prod_{k=n+1}^{2n}(n+1)^{1/(2n)}=\sqrt{n+1}\stackrel{n\to\infty}{\rightarrow}\infty.$$


TITLE: Is there a compact subset of the irrationals with positive Lebesgue measure?
QUESTION [12 upvotes]: Does there exist $K \subseteq \mathbb{R} \backslash \mathbb{Q}$ such that $K$ is compact, and has Lebesgue measure greater than $0$? As I have been trying to think of examples, I suspect that any subset of $\mathbb{R} \backslash \mathbb{Q}$ that is closed can be at most countable, since the closure of an uncountable subset of irrationals should contain some rationals. And, the Lebesgue measure of a countable set is $0$. If there are any examples of such a set, I would be very interested to know how it is constructed.

REPLY [18 votes]: The answer is yes.  Count the rationals in $[0,1]$ as $r_1,r_2,\ldots$, let $I_k$ be an open interval containing $r_k$ of length $3^{-k}$, and let $K=[0,1]\setminus\cup_k I_k$.
This question is somewhat related to the question Perfect set without rationals, but there measure did not come up.  For example, the Cantor set-like construction given there by JDH could be made to have positive measure.


TITLE: How to remember these probability results?
QUESTION [1 upvotes]: If $A,B$ and $C$ are  $3$ events, then

$P$(Exactly one of $A,B,C$ occurs)$=P(A)+P(B)+P(C)-2[P(A \cap B)+P(B \cap C)+P(A \cap C)]+3P(A \cap B \cap C)$
$P$(Exactly two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-3P(A \cap B \cap C)$
$P$(At least two of $A,B,C$ occur)$=P(A \cap B)+P(B \cap C)+P(A \cap C)-2P(A \cap B \cap C)$

REPLY [3 votes]: A picture is worth a thousand words


TITLE: Proof of Fourier Inversion Theorem in $\mathbb{R}^N$
QUESTION [0 upvotes]: As part of a proof of the Fourier Inversion Theorem in $\mathbb{R}^N$ I want to show that $$\lim_{\epsilon\to \infty}\frac{1}{(2\pi)^{N/2}\epsilon^N}\int_{\mathbb{R}^N} e^{-\frac{|x/\epsilon|^2}{2}} f(x) dL^N(x) = f(0).$$
My convention for the Fourier inversion is $\hat{f}(\xi) = \frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N} f(x)e^{-i\xi\cdot x} dL^N(x) $.
$f$ is assumed continuous at $x = 0$
I honestly don't know where to start. I feel like I want to pull the $\lim$ into the integral sign (using some convergence theorem perhaps?) and then maybe look at the expression $$\lim_{\epsilon \to \infty} \frac{e^{-i\xi\cdot x}}{\epsilon^N}.$$

REPLY [2 votes]: Start with changing the variable in the integral $x=\epsilon y$. $f(x)$ will be replaced by $f(\epsilon y)$, which will pointwise tend to $f(0)$ by continuity of $f$, the Jacobian will take care of the $\epsilon^{-N}$, a suitable convergence theorem will be invoked (e.g. Lebesgue's dominated convergence theroem) and everything will be OK.


TITLE: Ratio between inner products on a vector space
QUESTION [3 upvotes]: Let $V$ be a vector space and let$ \langle,\rangle_1$  and $\ \langle,\rangle_2$ be inner products in $V$ s.t. $\langle,\rangle_1=0  \iff \langle,\rangle_2=0$.
Prove $ \langle v,w\rangle_1=c\langle v,w\rangle_2$ for every $v,w \in V$.

I've been struggling coming up with a solution for this. This is what I have so far:
Let $U$ be a subspace of $V$. We know $\ U \oplus U^{\bot}=V$ for $\langle,\rangle_1$  and $\langle,\rangle_2$.
Let $\ B=\{u_1,...u_n\}$ be an orthonormal basis for $U$ and $\ C=\{v_1,...v_k\}$ an orthonormal basis for $\ U^{\bot}$ under $\langle,\rangle_1$.
$\ B'=\{\frac{u_1}{||u_1||},...,\frac{u_n}{||u_n||}\}$ is an orthonormal basis for $U$ and $\ C'=\{\frac{v_1}{||v_1||},...,\frac{v_k}{||v_k||}\}$ an orthonormal basis for $\ U^{\bot}$ under $\langle,\rangle_2$.
Take$\ u \in B $ and $\ v \in C$. If I prove for base vectors the argument is valid for any vector in the subspace.
I'd like to find the ratio using the norms of the bases B' and C', I'm not sure how to get at it and if this is the right direction.

REPLY [0 votes]: You do not need to assume finite dimensionality of $V$. The result you are trying to prove holds in any inner product space, real or complex, of any (possibly infinite) dimension. Summoning an orthonormal basis just complicates the problem. You should start by figuring out what the number $c$ must be. The answer is $c=\left\langle{u,u}\right\rangle_1/\left\langle{u,u}\right\rangle_2$, where $u$ is any nonzero vector in $V$ (If $V$ is the zero space, the statement to be proved is trivially true, so we can assume $V$ has nonzero vectors). For this definition of $c$ to make sense however, you have to show that it does not depend on the choice of $u$. In other words, you have to show that $$\frac{\left\langle{u,u}\right\rangle_1}{\left\langle{u,u}\right\rangle_2}=\frac{\left\langle{v,v}\right\rangle_1}{\left\langle{v,v}\right\rangle_2}$$ holds for all nonzero $u,v\in V$. Once you've done that, showing that $c$ is the desired ratio is straightforward.


TITLE: Minimum of functional $x\mapsto\langle Ax,x\rangle -\langle b,x\rangle$ when $A$ is not symmetric
QUESTION [0 upvotes]: In optimisation, a classic problem is the following:

Given positive definite matrix $A$ and a vector $b \in\Bbb{R}^n$, find the minimum of the functional $$x\mapsto\langle Ax,x\rangle -\langle b,x\rangle$$ in order to find a solution for the equation $Ax=b$.

I was wondering how this is useful in real life, since being positive definite seems to be a pretty strong condition. There must be some way this is generalized to the case where $A$ is just invertible, but I don't see how.

REPLY [2 votes]: If $A$ is invertible but not positive definite, it has an eigenvalue $\lambda < 0$. If $v$ is the corresponding eigenvector, then $\lim_{c \to \infty} (cv)^T A (cv) + b^T(cv) = \lim_{c \to \infty} \lambda c^2 + c (b^Tv) = -\infty$.
If $A$ is not symmetric, you can replace it with $0.5(A+A^T)$ and get the same function values (and same optimum).


TITLE: Number ways to distribute $k$ identical objects between $n$ people so that a special group of $a$ people receive odd number of objects among them
QUESTION [0 upvotes]: Suppose there is a group of $n$ people, and we need to choose exactly $k$ among them to give $k$ identical prizes. Also, exactly $a$ of these $n$ people are special. The problem is to calculate, how many ways are there to choose the $k$ people in such a way that, an odd number of special people are included in the choice?
For example, if $n = 5,\ k = 3,\ a = 3$ then there are $4$ possible ways. WLOG, we assume that the first $3$ people $p_1,\ p_2,\ p_3$ are special. Then the possible ways are:
$$
(p_1, p_2, p_3)\\
(p_1, p_4, p_5)\\
(p_2, p_4, p_5)\\
(p_3, p_4, p_5)
$$
I know the solution to a similar problem where the constraint of odd special people are not present, so we can pick any number of people from the $a$ special ones and the remaining from the $n - a$ non-special ones. There is bijection argument which shows the answer is simply ${n \choose k}$. But I can't seem to find any way to adapt this solution for the above problem with the odd constraint.

REPLY [0 votes]: The expression:
$$
\sum_{i=0}^{\left\lfloor\frac{a-1}2\right\rfloor}\binom{a}{2i+1}\binom{n-a}{k-2i-1}
$$
hardly can be generally simplified.


TITLE: How can I find eigenvalues and eigenvectors of matrix without computation
QUESTION [1 upvotes]: A is
$$
        \begin{pmatrix}
        0 & 1 & 0 & 0 \\
        0 & 0 & 1 & 0 \\
        0 & 0 & 0 & 1 \\
        0 & 0 & 0 & 0 \\
        \end{pmatrix}
$$
I already know that sum of eigenvalues is tr(A)=0 and times of eigenvlaues is det(A)=0. but how to determine eigenvalues??? and how can I get eigenvectors?

REPLY [1 votes]: Since it is an upper triangular matrix, so diagonal elements are eigenvalues. For eigenvalues vector at 0 , it can be easily seen that 2nd,3rd and 4th component is zero. So $X = (1,0,0,0)'$


TITLE: Will a rotating object traveling along a linear path continue to rotate forever in space?
QUESTION [2 upvotes]: Suppose engineers built a cylinder-shaped space ship which rotates to create artificial gravity:
  __________________
 /<-b \             \
/      \             \  -----> a
\      /             /
 \_b->/_____________/

The space ship uses a rocket to travel through space in the direction of a. Smaller nozzles ensure that the space ship rotates in in the direction indicated by b.

If the space ship wants to travel in a straight line, without stopping, would the astronauts need to frequently turn on the smaller nozzles to keep the space ship rotating at the proper speed or would the space ship constantly rotate at the same speed forever?
If the space ship was constantly rotated counter-clockwise to create artificial gravity, would this rotation cause the space ship to drift off course?

REPLY [6 votes]: There are two concepts here that may be getting mixed together, namely, conservation of linear momentum and conservation of angular momentum.
Newtons laws state that an object in motion will say in motion unless acted upon by an external force. So unless interstellar friction is a problem, the spaceship will keep travelling linearly  in the same direction forever (neglecting gravitational attraction to other bodies etc).
The same applies to rotational motion. A rotating object will continue to rotate unless acted upon by an external torque. So again, unless interstellar friction is a problem, the spaceship will continue to rotate forever.
To stress again, these two things are independent - one will not affect the other.

REPLY [1 votes]: If there is no torque acting on the ship, then its rotational velocity would stay constant.  The rotation of the ship has no effect on its course.


TITLE: Adjoint representation for matrix groups (Gauge theory)
QUESTION [1 upvotes]: This is a question in regards to an identity in Gauge theory.
Let $\omega$ be the connection one form on a principal bundle $\pi:P\rightarrow M$ and let $A_{\alpha}:=s_{\alpha}^*\omega$ be the gauge fields. ($s_{\alpha}$ is the local section associated to the local trivialisation).
On $U_{\alpha}\cap U_{\beta}\neq \varnothing$, there holds
\begin{align*}
A_{\alpha}&=\text{Ad}_{g_{\alpha\beta}}\circ(A_{\beta}-g_{\alpha\beta}^*\theta)
\end{align*}
For matrix groups, this equation becomes:
\begin{align*}
A_{\alpha}&=g_{\alpha\beta}A_{\beta}g_{\alpha\beta}^{-1}-dg_{\alpha\beta}g_{\alpha\beta}^{-1}.
\end{align*}
My question - can someone please explain to me why the top equation reduces to the bottom equation when considering matrix groups?
Thanks!

REPLY [1 votes]: In a matrix group, $Ad_g(A)=\displaystyle\frac{d}{dt}\bigg|_{t=0}g\exp(tA)g^{-1}=g(\displaystyle\frac{d}{dt}\bigg|_{t=0}\exp(tA))g^{-1}=gAg^{-1}$, i.e. the adjoint action, which is the derivative of conjugation in the Lie group, is really just conjugation in the larger space of all matrices. That's where the first term comes from.
From the context, I gather that $\theta$ is the Maurer-Cartan form. Let $X\in TM$. We have to compute
$Ad_{g_{\alpha\beta}}((g_{\alpha\beta})^*\theta(X))=Ad_{g_{\alpha\beta}}(\theta((g_{\alpha\beta})_*X)))=(R_{g_{\alpha\beta}^{-1}})_*(L_{g_{\alpha\beta}})_*(L_{g_{\alpha\beta}^{-1}})_*(g_{\alpha\beta})_*(X)=(R_{g_{\alpha\beta}^{-1}})_*(g_{\alpha\beta})_*(X)=(g_{\alpha\beta})_*(X)g_{\alpha\beta}^{-1}$.
Now, what is $(g_{\alpha\beta})_*(X)$? The function $g_{\alpha\beta}(x)$ is matrix-valued, so it has components $(a_{ij}(x))$ which are real-valued functions. Take some curve $\gamma(t)$ in $M$ with $\dot\gamma(0)=X$. Then
$(g_{\alpha\beta})_*(X)=\displaystyle\frac{d}{dt}\bigg|_{t=0}g_{\alpha\beta}(\gamma(t))=\displaystyle\frac{d}{dt}\bigg|_{t=0}(a_{ij}(\alpha(t)))=((a_{ij})_*(X))=(da_{ij}(X))$.
This explains the notation $dg_{\alpha\beta}$, which at first glance seems to make little sense. So by $dg_{\alpha\beta}$ is meant the matrix of one-forms $(da_{ij})$, and thus our computation in total is
$Ad_{g_{\alpha\beta}}((g_{\alpha\beta})^*\theta(X))=dg_{\alpha\beta}(X)g_{\alpha\beta}^{-1}$
which takes care of the second term.


text:
\begin{document}

\title{CONSTRUCTIONS OF $k$-g-FUSION FRAMES AND THEIR DUALS IN HILBERT SPACES}

\author[V. Sadri]{Vahid Sadri}
\address{Institute of Fundamental Sciences\\ University of Tabriz\\ Iran\\ Tel. 00989144112669.}
\email{vahidsadri57@gmail.com}

\author[R. Ahmadi]{Reza Ahmadi}
\address{Institute of Fundamental Sciences\\ University of Tabriz\\ Iran\\}
\email{rahmadi@tabrizu.ac.ir}

\author[A. Rahimi]{Asghar Rahimi}
\address{Department of Mathematics\\ University of Maragheh\\ Iran\\}
\email{rahimi@maragheh.ac.ir}

\begin{abstract}
Frames for operators or $k$-frames were recently considered by
G$\breve{\mbox{a}}$vruta (2012) in connection with atomic systems.
Also generalized frames are important frames in the Hilbert space of
bounded linear operators. Fusion frames, which are a special case of
generalized frames have various applications. This paper introduces
the concept of generalized fusion frames for operators aka
$k$-g-fusion frames and we get some results for characterization of
these frames. We further discuss on duals and Q-duals in connection
with $k$-g-fusion frames. Also we obtain some useful identities for
these frames. We also give several methods to construct $k$-g-fusion
frames. The results of this paper can be used in sampling theory
which are developed by g-frames and especially fusion frames. In the
end, we discuss the stability of a more general perturbation for
$k$-g-fusion frames.
\end{abstract}

\subjclass[2010]{Primary 42C15; Secondary 46C99, 41A58}

\keywords{Fusion frame, g-fusion frame, Dual  g-fusion frame, $k$-g-fusion, $Q$-dual $k$-g-fusion frame.}

\maketitle

\section{Introduction}
Frames in Hilbert spaces were first proposed by Duffin and Schaeffer
in the context of non-harmonic Fourier series \cite{ds}. Now, frames
have been widely applied in signal processing, sampling, filter bank
theory, system modeling, Quantum information, cryptography, etc
(\cite{bhf}, \cite{eldar}, \cite{fer}, \cite{st} and \cite{rens}).
We can say that fusion frames are the generalization of conventional
classical frames and special cases of g-frames in the field of frame
theory. The fusion frames are in fact more susceptible due to
complicated relations between the structure of the sequence of
weighted subspaces and the local frames in the subspaces and due to
the extreme sensitivity with respect to changes of the weights.

 Frames for operators or $k$-frames have been introduced by G$\breve{\mbox{a}}$vruta in \cite{ga} to study the nature of atomic systems for a separable Hilbert space  with respect to a bounded linear operator $k$. It is a well-known fact that $k$-frames are more general than the classical frames and due to higher generality of $k$-frames, many properties of frames may not hold for $k$-frames. Recently, we  presented g-fusion frames in \cite{sad}. This paper presents $k$-g-fusion frames with respect to a bounded linear operator on a separable Hilbert space which are a generalization of g-fusion frames.

Throughout this paper, $H$ and $K$ are separable Hilbert spaces and $\mathcal{B}(H,K)$ is the collection of all bounded linear operators of $H$ into $K$. If $K=H$, then $\mathcal{B}(H,H)$ will be denoted by $\mathcal{B}(H)$. Also, $\pi_{V}$ is the orthogonal projection from $H$ onto a closed subspace $V\subset H$ and  $\lbrace H_j\rbrace_{j\in\Bbb J}$ is a sequence of Hilbert spaces where $\Bbb J$ is a subset of $\Bbb Z$.

For the proof of the following lemma, refer to \cite{ga}.
\begin{lemma}\label{l1}
Let $V\subseteq H$ be a closed subspace, and $T$ be a linear  bounded operator on $H$. Then
$$\pi_{V}T^*=\pi_{V}T^* \pi_{\overline{TV}}.$$
If $T$ is a unitary (i.e. $T^*T=Id_{H}$), then
$$\pi_{\overline{Tv}}T=T\pi_{v}.$$
\end{lemma}
\begin{definition}\textbf{($k$-frame)}.
Let $\{f_j\}_{j\in\Bbb J}$ be a sequence of members of $H$ and
$k\in\mathcal{B}(H)$. We say that $\{f_j\}_{j\in\Bbb J}$ is a
$k$-frame  for $H$ if there exist $0<A\leq B<\infty$ such that for
each $f\in H$,
\begin{eqnarray*}
A\Vert k^*f\Vert^2\leq\sum_{j\in\Bbb J}\vert\langle f,f_j\rangle\vert^2\leq B\Vert f\Vert^2.
\end{eqnarray*}
\end{definition}
\begin{definition}(\textbf{g-fusion frame}).
Let $W=\lbrace W_j\rbrace_{j\in\Bbb J}$ be a collection of closed
subspaces of $H$, $\lbrace v_j\rbrace_{j\in\Bbb J}$ be a family of
weights, i.e. $v_j>0$  and $\Lambda_j\in\mathcal{B}(H,H_j)$ for each
$j\in\Bbb J$. We say $\Lambda:=(W_j, \Lambda_j, v_j)$ is a
generalized fusion frame (or g-fusion frame) for $H$ if there exist
$0<A\leq B<\infty$ such that for each $f\in H$,
\begin{eqnarray}\label{g}
A\Vert f\Vert^2\leq\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2\leq B\Vert f\Vert^2.
\end{eqnarray}
\end{definition}
If an operator $ u$ has closed range, then there exists a
right-inverse operator $u^ \dagger$ (pseudo-inverse of $u$) in the
following sences (see \cite{ch}).
\begin{lemma}\label{l3}
Let $u\in\mathcal{B}(K,H)$  be a bounded operator with closed range $\mathcal{R}_{u}$. Then there exists a bounded operator $u^\dagger \in\mathcal{B}(H,K)$ for which
$$uu^{\dagger} x=x, \ \ x\in \mathcal{R}_{u}.$$
\end{lemma}
\begin{lemma}\label{Ru}
Let $u\in\mathcal{B}(K,H)$. Then the following assertions holds:
\begin{enumerate} \item
 $\mathcal{R}_u$ is closed in $H$ if and only if $\mathcal{R}_{u^{\ast}}$ is closed in $K$.
\item $(u^{\ast})^\dagger=(u^\dagger)^\ast$.
\item
The orthogonal projection of $H$ onto $\mathcal{R}_{u}$ is given by $uu^{\dagger}$.
\item
The orthogonal projection of $K$ onto $\mathcal{R}_{u^{\dagger}}$ is given by $u^{\dagger}u$.\item$\mathcal{N}_{{u}^{\dagger}}=\mathcal{R}^{\bot}_{u}$ and $\mathcal{R}_{u^{\dagger}}=\mathcal{N}^{\bot}_{u}$.
 \end{enumerate}
\end{lemma}
\begin{lemma}(\cite{dag}).\label{dag}
Let $L_1\in\mathcal{B}(H_1,H)$ and $L_2\in\mathcal{B}(H_2,H)$ be  on given Hilbert spaces. Then the following assertions are equivalent:
\begin{enumerate}\item
$\mathcal{R}(L_1)\subseteq\mathcal{R}(L_2)$;\item
$L_1L_1^*\leq\lambda^2 L_2L_2^*$ for some$\lambda>0$;\item
there exists a  mapping $u\in\mathcal{B}(H,H_2)$ such that $L_1=L_2 u$.
\end{enumerate}

Moreover, if those condition are valid, then there exists a unique operator $u$ such that
\begin{enumerate}
\item[(a)] $\Vert u\Vert^2=\inf\{\alpha>0 \ \vert \ L_1L_1^*\leq\alpha L_2L_2^*\};$
\item[(b)] $\ker{L_1}=\ker{u};$
\item[(c)] $\mathcal{R}(u)\subseteq\overline{\mathcal{R}(L_2^*)}.$
\end{enumerate}
\end{lemma}

\section{$k$-g- Fusion Frames}
 We define the space $\mathscr{H}_2:=(\sum_{j\in\Bbb J}\oplus H_j)_{\ell_2}$ by
\begin{eqnarray}
\mathscr{H}_2=\big\lbrace \lbrace f_j\rbrace_{j\in\Bbb J} \ : \
f_j\in H_j , \ \sum_{j\in\Bbb J}\Vert f_j\Vert^2<\infty\big\rbrace
\end{eqnarray}
with the inner product defined by
$$\langle \lbrace f_j\rbrace, \lbrace g_j\rbrace\rangle=\sum_{j\in\Bbb J}\langle f_j, g_j\rangle.$$
It is clear that $\mathscr{H}_2$ is a Hilbert space with pointwise operations.
\begin{definition}
Let $W=\lbrace W_j\rbrace_{j\in\Bbb J}$ be a collection of closed
subspaces of $H$, $\lbrace v_j\rbrace_{j\in\Bbb J}$ be a family of
weights, i.e. $v_j>0$, $\Lambda_j\in\mathcal{B}(H,H_j)$ for each
$j\in\Bbb J$ and $k\in\mathcal{B}(H)$. We say $\Lambda:=(W_j,\Lambda_j, v_j)$ is a $k$-g- fusion frame  for $H$ if there exist
$0<A\leq B<\infty$  such that for each $f\in H$,
\begin{eqnarray}\label{g}
A\Vert k^*f\Vert^2\leq\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2\leq B\Vert f\Vert^2.
\end{eqnarray}
\end{definition}
When $k=id_H$, we get the g-fusion frame for $H$. Throughout this paper, $\Lambda$ will be a triple $(W_j, \Lambda_j, v_j)$ with $j\in\Bbb J$ unless otherwise noted.
We say $\Lambda$ is a Parseval $k$-g-fusion frame whenever
$$\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2=\Vert k^*f\Vert^2.$$
The synthesis and the analysis  operators in the $k$-g-fusion frames are defined by
\begin{align*}
T_{\Lambda}&:\mathscr{H}_2\longrightarrow H \ \ \ \ \ \ \ \ \ , \ \ \ \ \ \ T_{\Lambda}^*:H\longrightarrow\mathscr{H}_2\\
T_{\Lambda}(\lbrace f_j\rbrace_{j\in\Bbb J})&=\sum_{j\in\Bbb J}v_j \pi_{W_j}\Lambda_{j}^{*}f_j \ \ \ , \ \ \ \ T_{\Lambda}^*(f)=\lbrace v_j \Lambda_j \pi_{W_j}f\rbrace_{j\in\Bbb J}.
\end{align*}
Thus, the $k$-g-fusion frame operator is given by
$$S_{\Lambda}f=T_{\Lambda}T^*_{\Lambda}f=\sum_{j\in\Bbb J}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f$$
and
\begin{equation}\label{sf1}
\langle S_{\Lambda}f, f\rangle=\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2,
\end{equation}
for all $f\in H$. Therefore,
\begin{equation}\label{sf2}
\langle Akk^*f, f\rangle\leq \langle S_{\Lambda}f, f\rangle\leq \langle B f, f\rangle
\end{equation}
or
\begin{equation}\label{sf3}
Akk^*\leq  S_{\Lambda}\leq B I.
\end{equation}
Hence, we conclude that:
\begin{proposition}\label{pr3}
Let $\Lambda$ be a g-fusion Bessel sequence for $H$. Then $\Lambda$ is a $k$-g-fusion frame for $H$ if and only if there exists $A>0$ such that $S_{\Lambda}\geq Akk^*$.
\end{proposition}
\begin{remark}\label{re1}
In $k$-g-fusion frames, like $k$-frames and $k$-fusion frames, the $k$-g-fusion operator is not invertible. But,
if $k\in\mathcal{B}(H)$ has closed range, then the operator $S_\Lambda$ is an invertible operator on a subspace of $\mathcal{R}(k)\subset H$.
Indeed, suppose that $f\in\mathcal{R}(k)$, then
$$\Vert f\Vert^2=\Vert (k^\dagger\vert_{\mathcal{R}(k)})^*k^*f\Vert^2\leq\Vert k^\dagger\Vert^2 \Vert k^*f\Vert^2.$$
Thus, we have
$$A\Vert k^{\dagger}\Vert^{-2}\Vert f\Vert^2\leq\langle S_{\Lambda}f, f\rangle\leq B\Vert f\Vert^2,$$
which implies that $S_{\Lambda}:\mathcal{R}(k)\rightarrow S_{\Lambda}(\mathcal{R}(k))$ is a homeomorphism, furthermore, for each $f\in S_{\Lambda}(\mathcal{R}(k))$ we have
\begin{equation*}
B^{-1}\Vert f\Vert^2\leq\langle (S_{\Lambda}\vert_{\mathcal{R}(k)})^{-1}f, f\rangle\leq A^{-1}\Vert k^{\dagger}\Vert^2\Vert f\Vert^2.
\end{equation*}
\end{remark}
\begin{remark}\label{re2}
Since $S_{\Lambda}\in\mathcal{B}(H)$ is positive and self-adjoint and $\mathcal{B}(H)$ is a $C^*$-algabra, then $S^{-1}_{\Lambda}$ is positive and self-adjoint too whenever $k\in\mathcal{B}(H)$ has closed range. Now, for each $f\in S_{\Lambda}(\mathcal{R}(k))$ we can write
\begin{align*}
\langle kf, f\rangle&=\langle kf, S_{\Lambda}S_{\Lambda}^{-1}f\rangle\\
&=\langle S_{\Lambda}(kf), S_{\Lambda}^{-1}f\rangle\\
&=\langle\sum_{j\in\Bbb J}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}kf, S_{\Lambda}^{-1}f\rangle\\
&=\sum_{j\in\Bbb J}v_j^2 \langle S_{\Lambda}^{-1}\pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}kf, f\rangle.
\end{align*}
\end{remark}
\begin{theorem}
Let $u\in\mathcal{B}(H)$ be an invertible operator on $H$ and $\Lambda$ be a $k$-g-fusion frame for $H$ with bounds $A$ and $B$. Then, $\Gamma:=(uW_j, \Lambda_j \pi_{W_j}u^{*}, v_j)$ is a $uk$-g-fusion frame for $H$.
\end{theorem}
\begin{proof}
Let $f\in H$. Then by applying Lemma \ref{l1}, with $u$ instead of $T$, we have
\begin{align*}
\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}u^{*}\pi_{uW_j}f\Vert^2&=\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}u^{*}f\Vert^2\\
&\leq B\Vert u^{*}f\Vert^2\\
&\leq B\Vert u\Vert^2\Vert f\Vert^2.
\end{align*}
So, $\Gamma$ is a g-fusion Bessel sequence for $H$. On the other hand,
\begin{align*}
\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}u^{*}\pi_{uW_j}f\Vert^2&=\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j \pi_{W_j}u^{*}f\Vert^2\\
&\geq A\Vert k^*u^{*}f\Vert^2
\end{align*}
and the proof is completed.
\end{proof}
\begin{corollary}
If $u\in\mathcal{B}(H)$ is an invertible operator on Hilbert spaces, $\Lambda$ is a $k$-g-fusion frame for $H$ with bounds $A, B$ and $ku=uk$, then $\Gamma:=(uW_j, \Lambda_j \pi_{W_j}u^{*}, v_j)$ is a $k$-g-fusion frame for $H$.
\end{corollary}
\begin{theorem}
Let $u\in\mathcal{B}(H)$ be an invertible and unitary operator on $H$ and $\Lambda$ be a $k$-g-fusion frame for $H$ with bounds $A$ and $B$. Then, $(uW_j, \Lambda_j u^{-1}, v_j)$ is a $(u^{-1})^*k$-g-fusion frame for $H$.
\end{theorem}
\begin{proof}
Using Lemma \ref{l1},  we can write for any $f\in H$,
\begin{align*}
A\Vert k^*u^{-1}f\Vert^2\leq\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j u^{-1}\pi_{uW_j}f\Vert^2\leq B\Vert u^{-1}\Vert^2\Vert f\Vert^2.
\end{align*}
\end{proof}
\begin{corollary}
If $u\in\mathcal{B}(H)$ is an invertible and unitary operator on Hilbert spaces, $\Lambda$ is a $k$-g-fusion frame for $H$ with bounds $A, B$ and $k^*u=uk^*$, then $(uW_j, \Lambda_j u^{-1}, v_j)$ is a $k$-g-fusion frame for $H$.
\end{corollary}
\begin{theorem}\label{th2}
If $u\in\mathcal{B}(H)$, $\Lambda$ is a $k$-g-fusion frame for $H$ with bounds $A, B$ and $\mathcal{R}(u)\subseteq\mathcal{R}(k)$, then $\Lambda$ is a $u$-g-fusion frame for $H$.
\end{theorem}
\begin{proof}
Via Lemma \ref{dag}, there exists $\lambda>0$ such that $uu^*\leq\lambda^2kk^*$. Thus, for each $f\in H$ we have
$$\Vert u^*f\Vert^2=\langle uu^*f, f\rangle\leq\lambda^2\langle kk^*f, f\rangle=\lambda^2\Vert k^*f\Vert^2.$$
It follows that
$$\frac{A}{\lambda^2}\Vert u^*f\Vert^2\leq\sum_{j\in\Bbb J}v_j^2\Vert \Lambda_j\pi_{W_j}f\Vert^2.$$
\end{proof}
In the following example is showed that the condition $\mathcal{R}(u)\subseteq\mathcal{R}(k)$ in Corollary \ref{th2} is necessary.
\begin{example}
Let $H=\Bbb R^3$ and $\{e_1, e_2, e_3\}$ be an orthonormal basis for $H$. We define two operators $k$ and $u$ on $H$ by
\begin{align*}
&ke_1=e_2, \ \ \ ke_2=e_3, \ \ \ ke_3=e_3;\\
&ue_1=0, \ \ \ ue_2=e_1, \ \ \ ue_3=e_2.
\end{align*}
Suppose that $W_j=H_j:=\mbox{span}\{e_j\}$ where $j=1,2,3$. Let
\begin{align*}
\Lambda_j f:=\langle f, e_j\rangle e_j=f_j e_j,
\end{align*}
where $f=(f_1, f_2, f_3)$ and $j=1,2,3$. It is clear that $(W_j,
\Lambda_j, 1)$ is a $k$-g-fusion frame for $H$ with bounds
$\frac{1}{2}$ and $1$, respectively. Assume that $(W_j, \Lambda_j,
1)$ is a $u$-g-fusion frame for $H$. Then, by Proposition \ref{pr3},
there exists $A>0$ auch that $S_{\Lambda}=kk^*\geq Auu^*$. So, by
Lemma \ref{dag}, $\mathcal{R}(u)\subseteq \mathcal{R}(k)$. But, this
is a contradiction with $\mathcal{R}(k)\nsubseteq \mathcal{R}(u)$,
since $e_1\in \mathcal{R}(u)$ but $e_1\notin \mathcal{R}(k)$.
\end{example}
\begin{theorem}
Let $\Lambda:=(W_j, \Lambda_j, v_j)$ and $\Theta:=(W_j, \Theta_j, v_j)$ be two g-fusion Bessel sequences for $H$ with bounds $B_1$ and $B_2$, respectively. Suppose that $T_{\Lambda}$ and $T_{\Theta}$ be their analysis operators such that $T_{\Theta}T^*_{\Lambda}=k^*$ where $k\in\mathcal{B}(H)$. Then, both $\Lambda$ and $\Theta$ are $k$-g-fusion frames.
\end{theorem}
\begin{proof}
For each $f\in H$ we have
\begin{align*}
\Vert k^*f\Vert^4&=\langle k^*f, k^*f\rangle^2\\
&=\langle T^*_{\Lambda}f, T^*_{\Theta}k^*f\rangle^2\\
&\leq\Vert T^*_{\Lambda}f\Vert^2 \Vert T^*_{\Theta}k^*f\Vert^2\\
&=\big(\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2\big)\big(\sum_{j\in\Bbb I}v_j^2\Vert \Theta_j \pi_{W_j}k^*f\Vert^2\big)\\
&\leq\big(\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2\big) B_2 \Vert k^*f\Vert^2,
\end{align*}
thus, $B_2^{-1}\Vert k^*f\Vert^2\leq\sum_{j\in\Bbb I}v_j^2\Vert \Lambda_j \pi_{W_j}f\Vert^2$. This means that $\Lambda$ is a $k$-g-fusion frame for $H$. Similarly, $\Theta$ is a $k$-g-fusion frame with the lower bound $B_1^{-1}$.
\end{proof}

\section{$Q$-Duality of $k$-g-Fusion Frames}
In this section, we shall define duality of $k$-g-fusion frames and present some properties of them.
\begin{definition}
Let $\Lambda=(W_j, \Lambda_j, v_j)$ be a $k$-g-fusion frame for $H$. A g-fusion Bessel sequence $\tilde{\Lambda}:=(\tilde{W_j}, \tilde{\Lambda_j}, \tilde{v_j})$ is called $Q$-dual $k$-g-fusion frame (or brevity $Qk$-gf dual) for $\Lambda$ if there exists a bounded linear operator $Q:\mathscr{H}_2\rightarrow\mathscr{\tilde{H}}_2$ such that
\begin{equation}
T_{\Lambda}Q^*T_{\tilde{\Lambda}}^*=k,
\end{equation}
where $\mathscr{\tilde{H}}_2=(\sum_{j\in\Bbb J}\oplus \tilde{H}_j)_{\ell_2}$.
\end{definition}
Like another $k$-frames, the following presents equivalent conditions of the duality.
\begin{proposition}
Let $\tilde{\Lambda}$ be a $Qk$-gf dual for $\Lambda$. The following conditions are equvalent:
\begin{enumerate}
\item
$T_{\Lambda}Q^*T^*_{\tilde{\Lambda}}=k$;\item
$T_{\tilde{\Lambda}}QT^*_{\Lambda}=k^*$;\item
for each $f, f'\in H$, we have
$$\langle kf, f'\rangle=\langle T^*_{\tilde{\Lambda}}f,QT^*_{\Lambda}f'\rangle=\langle Q^*T^*_{\tilde{\Lambda}}f,T^*_{\Lambda}f'\rangle.$$
\end{enumerate}
\end{proposition}
\begin{proof}
Straightforward.
\end{proof}
\begin{theorem}
If $\tilde{\Lambda}$ is a $Qk$-gf dual for $\Lambda$, then $\tilde{\Lambda}$ is a $k^*$-g-fusion frame for $H$.
\end{theorem}
\begin{proof}
Let $f\in H$ and $B$ be an upper bound of $\Lambda$. Therefore,
\begin{align*}
\Vert kf\Vert^4&=\vert\langle kf, kf\rangle\vert^2\\
&=\vert\langle T_{\Lambda}Q^*T^*_{\tilde{\Lambda}}f, kf\rangle\vert^2\\
&=\vert\langle T^*_{\tilde{\Lambda}}f, QT^*_{\Lambda}kf\rangle\vert^2\\
&\leq\Vert T^*_{\tilde{\Lambda}}f\Vert^2\Vert Q\Vert^2 B\Vert kf\Vert^2\\
&=\Vert Q\Vert^2 B\Vert kf\Vert^2\sum_{j\in\Bbb J}\tilde{v}_j^2\Vert\tilde{\Lambda}_j\pi_{\tilde{W}_j}f\Vert^2
\end{align*}
and by definition, this completes the proof.
\end{proof}
\begin{corollary}
Assume $C_{op}$ and $D_{op}$ are the optimal bounds of $\tilde{\Lambda}$, respectively. Then
\begin{eqnarray*}
C_{op}\geq B_{op}^{-1}\Vert Q\Vert^{-2} \ \ \ and \ \ \ D_{op}\geq A_{op}^{-1}\Vert Q\Vert^{-2}
\end{eqnarray*}
which $A_{op}$ and $B_{op}$ are the optimal bounds of $\Lambda$, respectively.
\end{corollary}
Suppose that $\Lambda$ be a $k$-g-fusion frame for $H$. Since $S_F\geq Akk^*$, then by Lemma \ref{dag}, there exists an opeartor $u\in\mathcal{B}\big(H, \mathscr{H}_2)$ such that
\begin{eqnarray}\label{r8}
T_{\Lambda} u=k.
\end{eqnarray}
Now, we define the $j$-th component of $uf$ by $u_jf=(uf)_j$ for each $f\in H$. It is clear that $u_j\in\mathcal{B}(H, H_j)$. By this operator, we can construct some $Qk$-gf duals for $\Lambda$.
\begin{theorem}
Let $\Lambda$ be a $k$-g-fusion frame for $H$. If $u$ be an operator as in (\ref{r8}) and $\tilde{W_j}:=u^*_juW_j$. Then $\tilde{\Lambda}:=(\tilde{W_j}, \Lambda_j, v_j)$ is a  $Qk$-gf dual for $\Lambda$.
\end{theorem}
\begin{proof}
Define the mapping
\begin{align*}
\Phi_0:\mathcal{R}(T_{\tilde{\Lambda}}^*)&\rightarrow\mathscr{H}_2,\\
\Phi_0(T_{\tilde{\Lambda}}^*f)&=uf.
\end{align*}
Then $\Phi_0$ is well-defined. indeed, if $f_1 ,f_2\in H$ and $T_{\tilde{\Lambda}}^*f_1=T_{\tilde{\Lambda}}^*f_2$, then $\pi_{\tilde{W_j}}(f_1-f_2)=0$. Therefore, for any $j\in\Bbb J$,
$$f_1-f_2\in(\tilde{W_j})^{\perp}=\mathcal{R}(u_j^*)^{\perp}=\ker{u_j}.$$
Thus, $uh_1=uh_2$.
It is clear that $\Phi_0$ be bounded and linear. Therefore, it has a unique linear extension (also denoted $\Phi_0$) to $\overline{\mathcal{R}(T_{\tilde{\Lambda}}^*)}$. Define $\Phi$ on $\mathscr{H}_2$ by setting
\begin{align*}
\Phi=\begin{cases}
\Phi_0,& \mbox{on} \ \overline{\mathcal{R}(T_{\tilde{\Lambda}}^*)},\\
0,& \mbox{on} \ \overline{\mathcal{R}(T_{\tilde{\Lambda}}^*)}^{\perp}
\end{cases}
\end{align*}
and let $Q:=\Phi^*$. This implies that $Q^*\in\mathcal{B}(\mathscr{H}_2, \mathscr{H}_2)$ and
$$T_{\Lambda}Q^*T_{\tilde{\Lambda}}^*=T_{\Lambda}\Phi T_{\tilde{\Lambda}}^*=T_{\Lambda} u=k.$$
\end{proof}
\begin{proposition}
Let $\Lambda$ be a $k$-g-fusion frame with optimal bounds of $A_{op}$ and $B_{op}$, respectively and $k$ has closed range. Then
\begin{eqnarray*}
B_{op}=\Vert S_{\Lambda}\Vert=\Vert T_{\Lambda}\Vert^2 \ \ \ , \ \ \ A_{op}=\Vert u_0\Vert^{-2}
\end{eqnarray*}
which $u_0$ is the unique solotion of equation (\ref{r8}).
\end{proposition}
\begin{proof}
Via Lemma \ref{dag}, the equation (\ref{r8}) has a unique solition as $u_0$ such that
\begin{align*}
\Vert u_0\Vert^2&=\inf\{\alpha>0 \ \vert \ kk^*\leq\alpha T_{\Lambda}T^{*}_{\Lambda}\}\\
&=\inf\{\alpha>0 \ \vert \ \Vert k^* f\Vert^2\leq\alpha \Vert T^*_{\Lambda} f\Vert^2 \ , \ f\in H\}.
\end{align*}
Now, we have
\begin{align*}
A_{op}&=\sup\{A>0 \ \vert \ A\Vert k^*f\Vert^2\leq\Vert T^*_{\Lambda} f\Vert^2 \ , \ f\in H\}\\
&=\Big(\inf\{\alpha>0 \ \vert \ \Vert k^*f\Vert^2\leq\alpha\Vert T^*_{\Lambda} f\Vert^2 \ , \ f\in H\}\Big)^{-1}\\
&=\Vert u_0\Vert^{-2}.
\end{align*}
\end{proof}
\subsection{$k$-g-fusion duals and some identities}
\begin{definition}\label{dual}
Let $\Lambda$ be a $k$-g-fusion frame for $H$. A g-fusion Bessel
sequence $\tilde{\Lambda}=(\tilde{W_j}, \tilde{\Lambda_j}, v_j)$ is
called a $k$-g-fusion dual of $\Lambda$ if for each $f\in H$,
\begin{equation}\label{d1}
kf=\sum_{j\in\Bbb J}v_j^2\pi_{W_j}\Lambda^*_{j}\tilde{\Lambda_j}\pi_{\tilde{W_j}}f.
\end{equation}
\end{definition}
In this case, we can deduce that $\tilde{\Lambda}$ is a $k^*$-g-fusion frame for $H$. Indeed, for each $f\in H$ we have
\begin{align*}
\Vert kf\Vert^4&\leq\big\vert\langle\sum_{j\in\Bbb J}v_j^2\pi_{W_j}\Lambda^*_{j}\tilde{\Lambda_j}\pi_{\tilde{W_j}}f, kf\rangle\big\vert^2\\
&=\big\vert\sum_{j\in\Bbb J}v_j^2\langle\tilde{\Lambda_j}\pi_{\tilde{W_j}}f, \Lambda_j\pi_{W_j}kf\rangle\big\vert^2\\
&\leq\Big(\sum_{j\in\Bbb J}v_j^2\Vert\tilde{\Lambda_j}\pi_{\tilde{W_j}}f\Vert^2\Big)\Big(\sum_{j\in\Bbb J}v_j^2\Vert\Lambda_j\pi_{W_j}kf\Vert^2\Big)\\
&\leq B\Vert kf\Vert^2\sum_{j\in\Bbb J}v_j^2\Vert\tilde{\Lambda_j}\pi_{\tilde{W_j}}f\Vert^2,
\end{align*}
where $B$ is an upper bound of $\Lambda$.
\begin{theorem}
Let $\Lambda$ be a $k$-g-fusion frame for $H$ with bounds $A, B$ and $k$ be closed range. Then $\Big(k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j, \Lambda_j\pi_{W_j}\pi_{S_{\Lambda}(\mathcal{R}(k))}(S_{\Lambda}^{-1})^*k, v_j\Big)$ is a $k$-dual of $\Lambda$.
\end{theorem}
\begin{proof}
We know that $S_{\Lambda}^{-1}S_{\Lambda}\vert_{\mathcal{R}(k)}=Id_{\mathcal{R}(k)}$. Then, we have for each $f\in H$,
\begin{align*}
kf&=S_{\Lambda}(S_{\Lambda}^{-1})^*kf\\
&=S_{\Lambda}\pi_{S_{\Lambda}(\mathcal{R}(k))}(S_{\Lambda}^{-1})^*kf\\
&=\sum_{j\in\Bbb J}v_j^2\pi_{W_j}\Lambda^*_j\Lambda_j\pi_{W_j}\pi_{S_{\Lambda}(\mathcal{R}(k))}(S_{\Lambda}^{-1})^*kf\\
&=\sum_{j\in\Bbb J}v_j^2\pi_{W_j}\Lambda^*_j\Lambda_j\pi_{W_j}\pi_{S_{\Lambda}(\mathcal{R}(k))}(S_{\Lambda}^{-1})^*k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f.
\end{align*}
On the other hand, we obtain by Remark \ref{re1}, for all $f\in H$,
\begin{small}
\begin{align*}
\sum_{j\in\Bbb J}v_j^2&\Vert\Lambda_j\pi_{W_j}\pi_{S_{\Lambda}(\mathcal{R}(k))}(S_{\Lambda}^{-1})^*k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f\Vert^2\\
&=\Big\langle S_{\Lambda}\big((S_{\Lambda}^{-1})^*k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f\big), (S_{\Lambda}^{-1})^*k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f\Big\rangle\\
&=\Big\langle k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f\big), (S_{\Lambda}^{-1})^*k\pi_{k^*S_{\Lambda}^{-1}\pi_{S_{\Lambda}(\mathcal{R}(k))}W_j}f\Big\rangle\\
&\leq A^{-1}\Vert k^{\dagger}\Vert^2\Vert k\Vert^2\Vert f\Vert^2
\end{align*}
\end{small}
and the proof is completed by Definition \ref{dual}.
\end{proof}
Let $\Lambda$ be a $k$-g-fusion frame for $H$ and $\tilde{\Lambda}$ be a $k$-g-fusion dual of $\Lambda$. Suppose that $\Bbb I$ is a finite subset of $\Bbb J$ and we define
\begin{align}
S_{\Bbb I}f=\sum_{j\in\Bbb I}v_j^2\pi_{W_j}\Lambda^*_j\tilde{\Lambda_j}\pi_{\tilde{W_j}}f , \ \ \ \ \ (\forall f\in H).
\end{align}
It is easy to check that $S_{\Bbb I}\in\mathcal{B}(H)$, positive and
$$S_{\Bbb I}+S_{\Bbb I^c}=k.$$
\begin{theorem}\label{tg1}
Let $f\in H$, then
\begin{eqnarray*}
\sum_{j\in\Bbb I}v_j^2\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}kf\rangle-\Vert S_{\Bbb I}f\Vert^2 =\sum_{j\in\Bbb I^c}v_j^2\overline{\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}kf\rangle}-\Vert S_{\Bbb I^c}f\Vert^2.
\end{eqnarray*}
\end{theorem}
\begin{proof}
For each $f\in H$, we have
\begin{align*}
\sum_{j\in\Bbb I}v_j^2\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}kf\rangle-\Vert S_{\Bbb I}f\Vert^2&=\langle k^*S_{\Bbb I}f,f\rangle-\Vert S_{\Bbb I}f\Vert^2\\
&=\langle k^*S_{\Bbb I}f,f\rangle-\langle S^{\ast}_{\Bbb I} S_{\Bbb I}f,f\rangle\\
&=\langle(k-S_{\Bbb I})^{\ast}S_{\Bbb I}f,f\rangle\\
&=\langle S^{\ast}_{\Bbb I^c}(k-S_{\Bbb I^c})f,f\rangle\\
&=\langle S^{\ast}_{\Bbb I^c}kf,f\rangle-\langle S^{\ast}_{\Bbb I^c}S_{\Bbb I^c}f,f\rangle\\
&=\langle f, k^*S_{\Bbb I^c}f\rangle-\langle S_{\Bbb I^c}f,S_{\Bbb I^c}f\rangle\\
&=\sum_{j\in\Bbb I^c}v_j^2\overline{\langle \tilde{\Lambda}_j \pi_{\tilde{W_j}}f,\Lambda_j \pi_{W_j}kf\rangle}-\Vert S_{\Bbb I^c}f\Vert^2
\end{align*}
and the proof is completed.
\end{proof}
\begin{theorem}\label{ti1}
Let $\Lambda$ be a Parseval $k$-g-fusion frame for $H$. If $\Bbb I\subseteq\Bbb J$ and  $E\subseteq\Bbb I^c$, then for each $f\in H$,
\begin{align*}
\Vert\sum_{j\in \Bbb I\cup E}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2&-\Vert\sum_{j\in\Bbb I^c\setminus E}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2\\
&=\Vert\sum_{j\in \Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2-\Vert\sum_{j\in\Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2\mbox{Re}\Big(\sum_{j\in E}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big).
\end{align*}
\end{theorem}
\begin{proof}
Let
$$S_{\Lambda, \Bbb I}f:=\sum_{j\in\Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f.$$
Therefore, $S_{\Lambda, \Bbb I}+S_{\Lambda, \Bbb I^c}=kk^*$. Hence,
\begin{align*}
S^2_{\Lambda, \Bbb I}-S^2_{\Lambda, \Bbb I^c}&=S^2_{\Lambda, \Bbb I}-(kk^*-S_{\Lambda, \Bbb I})^2\\
&=kk^*S_{\Lambda, \Bbb I}+S_{\Lambda, \Bbb I}kk^*-(kk^*)^2\\
&=kk^*S_{\Lambda, \Bbb I}-S_{\Lambda, \Bbb I^c}kk^*.
\end{align*}
Now, for each $f\in H$ we have
$$\Vert S^2_{\Lambda, \Bbb I}f\Vert^2-\Vert S^2_{\Lambda, \Bbb I^c}\Vert^2=\langle kk^*S_{\Lambda, \Bbb I}f, f\rangle-\langle S_{\Lambda, \Bbb I^c}kk^*f, f\rangle.$$
Consequently, for $\Bbb I\cup E$ instead of $\Bbb I$:
\begin{align*}
&\Vert\sum_{j\in \Bbb I\cup E}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2-\Vert\sum_{j\in\Bbb I^c\setminus E}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2\\
=&\sum_{j\in \Bbb I\cup E}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle-\sum_{j\in \Bbb I^c\setminus E}v_j^2\overline{\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle}\\
=&\sum_{j\in \Bbb I}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle-\sum_{j\in \Bbb I^c}v_j^2\overline{\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ +2\mbox{Re}\Big(\sum_{j\in E}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big)\\
=&\Vert\sum_{j\in \Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2-\Vert\sum_{j\in\Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2\mbox{Re}\Big(\sum_{j\in E}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big).
\end{align*}
\end{proof}
\begin{theorem}
Let $\Lambda$ be a Parseval $k$-g-fusion frame for $H$. If $\Bbb I\subseteq\Bbb J$ and  $E\subseteq\Bbb I^c$, then for any $f\in H$,
\begin{align*}
&\Vert\sum_{j\in \Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2+\mbox{Re}\Big(\sum_{j\in \Bbb I^c}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big)\\
=&\Vert\sum_{j\in \Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2+\mbox{Re}\Big(\sum_{j\in \Bbb I}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big)\geq\frac{3}{4}\Vert kk^*f\Vert^2.
\end{align*}
\end{theorem}
\begin{proof}
In Theorem \ref{ti1}, we showed that
\begin{align*}
S^2_{\Lambda, \Bbb I}-S^2_{\Lambda, \Bbb I^c}=kk^*S_{\Lambda, \Bbb I}-S_{\Lambda, \Bbb I^c}kk^*.
\end{align*}
Therefore,
$$S^2_{\Lambda, \Bbb I}+S^2_{\Lambda, \Bbb I^c}=2\Big(\frac{kk^*}{2}-S_{\Lambda, \Bbb I}\Big)^2+\frac{(kk^*)^2}{2}\geq\frac{(kk^*)^2}{2}.$$
Thus,
\begin{align*}
kk^*S_{\Lambda, \Bbb I}+S^2_{\Lambda, \Bbb I^c}+(kk^*S_{\Lambda, \Bbb I}+S^2_{\Lambda, \Bbb I^c})^*&=kk^*S_{\Lambda, \Bbb I}+S^2_{\Lambda, \Bbb I^c}+S_{\Lambda, \Bbb I}kk^*+S^2_{\Lambda, \Bbb I^c}\\
&=kk^*(S_{\Lambda, \Bbb I}+S_{\Lambda, \Bbb I^c})+S^2_{\Lambda, \Bbb I}+S^2_{\Lambda, \Bbb I^c}\\
&\geq\frac{3}{2}(kk^*)^2.
\end{align*}
Now, we obtain for any $f\in H$,
\begin{align*}
&\Vert\sum_{j\in \Bbb I}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2+\mbox{Re}\Big(\sum_{j\in \Bbb I^c}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big)\\
=&\Vert\sum_{j\in \Bbb I^c}v_j^2 \pi_{W_j}\Lambda^*_j \Lambda_j \pi_{W_j}f\Vert^2+\mbox{Re}\Big(\sum_{j\in \Bbb I}v_j^2\langle\Lambda_j\pi_{W_j}f, \Lambda_j\pi_{W_j}kk^*f\rangle\Big)\\
&=\frac{1}{2}\big(\langle kk^*S_{\Lambda, \Bbb I}f, f\rangle+\langle S^2_{\Lambda, \Bbb I^c}f, f\rangle+\langle f, kk^*S_{\Lambda, \Bbb I}f\rangle+\langle f, S^2_{\Lambda, \Bbb I^c}f\rangle\big)\\
&\geq\frac{3}{4}\Vert kk^*f\Vert^2.
\end{align*}
\end{proof}

\section{Perturbation of $k$-g-Fusion Frames}
Perturbation of frames has been discussed by Cazassa and Christensen in \cite{caz}. In this section, we present some perturbation of $k$-g-fusion frames and review some results about them.
\begin{lemma}(\cite{caz})\label{G1}
Let U be a Linear operator on a Banach space X and assume that there exist  $\lambda_{1},\lambda_{2}\in[0,1) $ such that
 \begin{align*}
  & \Vert x-Ux\Vert \leq\lambda_{1}\Vert x\Vert+\lambda_{2}\Vert Ux\Vert
\end{align*}
for all $ x\in X$. Then $U$ is bounded and invertible. Moreovere,
 \begin{align*}
 \frac{1-\lambda_{1}}{1+\lambda_{2}}\Vert x\Vert\leq\Vert Ux\Vert\leq\frac{1+\lambda_{1}}{1-\lambda_{2}}\Vert x\Vert
 \end{align*}
 and
 \begin{align*}
 \frac{1-\lambda_{2}}{1+\lambda_{1}}\Vert x\Vert\leq\Vert U^{-1}x\Vert\leq\frac{1+\lambda_{2}}{1-\lambda_{1}}\Vert x\Vert
 \end{align*}
 for all $x\in X$.
\end{lemma}
\begin{theorem}\label{p1}
Let $k\in\mathcal{B}(H)$  be closed range, $\Lambda$ be a $k$-g-fusion frame  for $H$ with bounds $A,B$ and $\lbrace\Theta_{j}\in\mathcal{B}(H, H_j)\rbrace_{j\in\Bbb J}$ be a sequence of operators  such that for any finite subset $\Bbb I\subseteq \Bbb J$ and for each $ f\in H$,
\begin{small}
\begin{align*}
\Vert \sum_{j\in\Bbb I}v_{j}^2\big (\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f-&\pi_{W_{j}}\Theta^*_{j}\Theta_{j}\pi_{W_{j}}f\big)\Vert \leq
\lambda_1 \Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Lambda^*_j\Lambda_j\pi_{W_{j}} f\Vert\\
& +\lambda_2 \Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Theta^*_j\Theta_j\pi_{W_{j}} f\Vert
 +\gamma (\sum_{j\in\Bbb I}v_{j}^2\Vert\Lambda_j\pi_{W_{j}} f\Vert^{2})^{\frac{1}{2}},
\end{align*}
\end{small}
where $ 0\leq \max\lbrace \lambda_1+\frac{\gamma}{\sqrt{A}}, \lambda_2\rbrace<1$, then $\Theta:=(W_j, \Theta_j, v_j)$ is a $k$-g-fusion frame for $H$ with bounds
$$ A\frac{1-(\lambda_1+\frac{\gamma}{\sqrt{A}})}{1+\lambda_2} \ \  and \ \ B\big(\frac{1+\lambda_1+\frac{\gamma}{\sqrt{B}}}{1-\lambda_2}\big)$$
\end{theorem}
\begin{proof}
Assume that  $\Bbb I\subseteq \Bbb J$ is a finite subset and  $f\in H$. We have
\begin{small}
\begin{align*}
\Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Theta^*_ j\Theta_ j\pi_{W_{j}}f\Vert &\leq \Vert \sum_{j\in\Bbb I}v_{j}^2\big (\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f-\pi_{W_{j}}\Theta^*_{j}\Theta_{j}\pi_{W_{j}}f\big)\Vert\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  +\Vert \sum_{j\in\Bbb I}v_{j}^2 \pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f\Vert\\
&\leq (1+\lambda_1)\Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f\Vert +\lambda_2\Vert \sum_{j\in\Bbb I }v_{j}^2\pi_{W_{j}}\Theta^*_ j\Theta_ j\pi_{W_{j}}f\Vert\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\gamma(\sum_{j\in\Bbb I}v_{j}^2\Vert\Lambda_j\pi_{W_{j}} f\Vert^{2})^{\frac{1}{2}},
\end{align*}
then,
\begin{align*}
\Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Theta^*_ j\Theta_ j\pi_{W_{j}}f\Vert&\leq \frac{1+\lambda_1}{1-\lambda_2}\Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f\Vert +\frac{\gamma}{1-\lambda_2} ( \sum_{j\in\Bbb I}v_{j}^2\Vert\Lambda_j\pi_{w_{j}}f\Vert^{2})^\frac{1}{2}\\
&\leq\Big(\frac{1+\lambda_1}{1-\lambda_2}B+\frac{\gamma}{1-\lambda_2}\sqrt{B}\Big)\Vert f\Vert<\infty.
\end{align*}
\end{small}
Therefore,
$\sum_{j\in\Bbb J} v_{j}^2\pi_{W_{j}}\Theta^*_ j\Theta_ j\pi_{W_{j}}f$
is unconditionally convergent.
It follows that $ \Theta:=(W_{j},\Theta_{j}, v_{j})$ is a g-fusion Bessel sequence for $H$ with bound
$$(\frac{1+\lambda_1}{1-\lambda_2}B+\frac{\gamma}{1-\lambda_2}\sqrt{B}).$$
Thus, we obtain by the hypothesis
\begin{align*}
\Vert S_{\Lambda}f-S_{\Theta}f\Vert\leq\lambda_1\Vert S_{\Lambda}f\Vert + \lambda_2\Vert S_{\Theta}f\Vert +\gamma (\sum_{j\in\Bbb I}v_{j}^2\Vert\Lambda_{j}\pi_{w_{j}}f\Vert)^2)^\frac{1}{2}.
\end{align*}
Therefore, by (\ref{sf1})
\begin{align*}
\Vert f-S_{\Theta}S_{\Lambda}^{-1}f\Vert&\leq\lambda_1\Vert f\Vert+\lambda_2\Vert S_{\Theta}S_{\Lambda}^{-1}f\Vert+\gamma\big(\sum_{j\in\Bbb J}v_J^2\Vert\Lambda_j\pi_{W_j}S_{\Lambda}^{-1}f\Vert^2\big)^{\frac{1}{2}}\\
&\leq\Big(\lambda_1+\frac{\gamma}{\sqrt{A}}\Big)\Vert f\Vert+\lambda_2\Vert S_{\Theta}S_{\Lambda}^{-1}f\Vert.
\end{align*}
Since $0\leq \max\lbrace \lambda_1+\frac{\gamma}{\sqrt{A}}, \lambda_2\rbrace<1$, then by Lemma \ref{G1},  $S_{\Theta}S_{\Lambda}^{-1}$ and consequently $S_{\Theta}$ is invertible and we get
$$\frac{1-\lambda_2}{1+\big(\lambda_1+\frac{\gamma}{\sqrt{A}}\big)}\leq \Vert S_{\Lambda}S_{\Theta}^{-1}\Vert\leq\frac{1+\lambda_2}{1+\big(\lambda_1-\frac{\gamma}{\sqrt{A}}\big)}.$$
So,
\begin{align*}
\Vert S_{\Theta}\Vert&\geq\frac{A}{\Vert S_{\Lambda}S_{\Theta}^{-1}\Vert}\\
&\geq\frac{A-(A\lambda_1+\gamma\sqrt{A})}{1+\lambda_2}\Vert kk^*\Vert
\end{align*}
and by Proposition \ref{pr3}, $\Theta$ is a $k$-g-fusion frame for $H$.
\end{proof}
\begin{corollary}\label{p2}
Let  $\Lambda$ be a $k$-g-fusion frame  for $H$ with bounds $A,B$ and $\lbrace\Theta_{j}\in\mathcal{B}(H, H_j)\rbrace_{j\in\Bbb J}$ be a sequence of operators. If there exists a constant $0<R<A$ such that
\begin{align*}
 \sum_{j\in\Bbb J}v_{j}^2\Vert\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f-\pi_{W_{j}}\Theta^*_{j}\Theta_{j}\pi_{W_{j}}f\Vert \leq R\Vert k^*f\Vert
\end{align*}
for all $f\in H$, then $\Theta:=(W_j, \Theta_j, v_j)$ is a g-fusion frame for $H$ with bounds
$$A-R \ \ \ \ and \ \ \ \ \min\big\lbrace B+R\sqrt{\frac{B}{A}}, R\Vert k\Vert+\sqrt{B}\big\rbrace.$$
\end{corollary}
\begin{proof}
It is easy to check that $\sum_{j\in\Bbb
J}v_j^2\pi_{W_{j}}\Theta^*_{j}\Theta_{j}\pi_{W_{j}}f$ is convergent
for any $f\in H$. Thus, we obtain for each $f\in H$,
\begin{align*}
\sum_{j\in\Bbb J}v_j^2\Vert \pi_{W_j}\Lambda^*_j\Lambda_j\pi_{W_j}f-\pi_{W_j}\Theta^*_j\Theta_j\pi_{W_j}f\Vert&\leq R\Vert k^*f\Vert\\
&\leq\frac{R}{\sqrt{A}}\big(\sum_{j\in\Bbb J}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2\big)^{\frac{1}{2}}
\end{align*}
and also
\begin{align*}
\sum_{j\in\Bbb J}v_j^2\Vert \pi_{W_j}\Theta^*_j\Theta_j\pi_{W_j}f\Vert&\leq R\Vert k^*f\Vert+\sum_{j\in\Bbb J}v_j^2\Vert\pi_{W_j}\Lambda^*_j\Lambda_j\pi_{W_j}f\Vert\\
&\leq(R\Vert k\Vert+\sqrt{B})\Vert f\Vert.
\end{align*}
By using Theorem \ref{p1} with  $\lambda_1=\lambda_2=0$ and $\gamma=\frac{R}{\sqrt{A}}$, the proof is completed.
\end{proof}
The proof of the following is similar to the proof of Theorem \ref{p1}.
\begin{theorem}
Let $\Lambda$ be a $k$-g-fusion frame for $H$ with bounds $A,B$ and $\lbrace\Theta_{j}\in\mathcal{B}(H, H_j)\rbrace_{j\in\Bbb J}$ be a sequence of operators  such that for any finite subset $\Bbb I\subseteq \Bbb J$ and for each $f\in H$,
\begin{small}
\begin{align*}
\Vert \sum_{j\in\Bbb I}v_{j}^2\big (\pi_{W_{j}}\Lambda^*_ j\Lambda_ j\pi_{W_{j}}f-&\pi_{W_{j}}\Theta^*_{j}\Theta_{j}\pi_{W_{j}}f\big)\Vert \leq
\lambda_1 \Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Lambda^*_j\Lambda_j\pi_{W_{j}} f\Vert\\
& +\lambda_2 \Vert \sum_{j\in\Bbb I}v_{j}^2\pi_{W_{j}}\Theta^*_j\Theta_j\pi_{W_{j}} f\Vert
 +\gamma \Vert k^*f\Vert,
\end{align*}
\end{small}
where $ 0\leq \max\lbrace \lambda_1+\frac{\gamma}{\sqrt{A}\Vert k\Vert}, \lambda_2\rbrace<1$, then $\Theta:=(W_j, \Theta_j, v_j)$ is a $k$-g-fusion frame for $H$ with bounds
$$ A\frac{1-(\lambda_1+\frac{\gamma}{\sqrt{A}}\Vert k\Vert)}{1+\lambda_2} \ \  and \ \ B\big(\frac{1+\lambda_1+\frac{\gamma}{\sqrt{B}}\Vert k\Vert}{1-\lambda_2}\big).$$
\end{theorem}
\begin{theorem}
Let  $\Lambda$ be a $k$-g-fusion frame  for $H$ with bounds $A,B$ and $\lbrace\Theta_{j}\in\mathcal{B}(H, H_j)\rbrace_{j\in\Bbb J}$ be a sequence of operators. If there exists a constant $0<R<A$ such that
\begin{align*}
 \sum_{j\in\Bbb J}v_{j}^2\Vert\Lambda_ j\pi_{W_{j}}f-\Theta_{j}\pi_{W_{j}}f\big)\Vert^2 \leq R\Vert k^*f\Vert^2
\end{align*}
for all $f\in H$, then $\Theta:=(W_j, \Theta_j, v_j)$ is a $k$-g-fusion frame for $H$ with bounds
$$(\sqrt{A}-\sqrt{R})^2 \ \ \ \ and \ \ \ \ (\Vert k\Vert\sqrt{R}+\sqrt{B})^2.$$
\end{theorem}
\begin{proof}
Let $f\in H$. By the triangle and Minkowski inequality, we can write
\begin{small}
\begin{align*}
\Big(\sum_{j\in\Bbb J}v_j^2\Vert\Theta_j\pi_{W_j}f\Vert^2\Big)^{\frac{1}{2}}&\leq
\Big(\sum_{j\in\Bbb J}v_{j}^2\Vert\Lambda_ j\pi_{W_{j}}f-\Theta_{j}\pi_{W_{j}}f\big)\Vert^2\Big)^{\frac{1}{2}}+\Big(\sum_{j\in\Bbb J}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2\Big)^{\frac{1}{2}}\\
&\leq(\Vert k\Vert\sqrt{R}+\sqrt{B})\Vert f\Vert.
\end{align*}
\end{small}
Also
\begin{small}
\begin{align*}
\Big(\sum_{j\in\Bbb J}v_j^2\Vert\Theta_j\pi_{W_j}f\Vert^2\Big)^{\frac{1}{2}}&\geq
\Big(\sum_{j\in\Bbb J}v_j^2\Vert\Lambda_j\pi_{W_j}f\Vert^2\Big)^{\frac{1}{2}}-
\Big(\sum_{j\in\Bbb J}v_{j}^2\Vert\Lambda_ j\pi_{W_{j}}f-\Theta_{j}\pi_{W_{j}}f\big)\Vert^2\Big)^{\frac{1}{2}}\\
&\geq(\sqrt{A}-\sqrt{R})\Vert k^*f\Vert.
\end{align*}
\end{small}
Thus, these complete the proof.
\end{proof}

config: arxiv


TITLE: Proving $(A - B) \times (C - D) = (A \times C - B \times C) - A \times D$
QUESTION [2 upvotes]: I am trying to prove $(A - B) \times (C - D) = (A \times C - B \times C) - A \times D$ using biconditionals, but cannot quite get there. For any ordered tuple $(\alpha, \beta)$, we have:
\begin{align*}
(\alpha, \beta) \in (A - B) \times (C - D) & \iff \alpha \in (A - B) \land \beta \in (C - D) \\
& \iff (\alpha \in A \land \alpha \not \in \beta) \land (\beta \in C \land \beta \not \in D) \\
& \iff ((\alpha \in A \land \alpha \not \in \beta) \land \beta \in C) \land (\beta \not \in D) \\
& \iff ((\alpha \in A \land \alpha \not \in \beta) \land \beta \in C) \land (\alpha \in A \land \beta \not \in D) \\
& \iff ((\alpha \in A \land \beta \in C) \land (\alpha \not \in B \land \beta \in C)) \land (\alpha \in A \land \beta \not \in D) \\
& \iff (((\alpha, \beta) \in A \times B) \land (\alpha \not \in B \land \beta \in C)) \land (\alpha, \beta) \not \in A - D.
\end{align*}
At this point, I do not know how to finish. $\alpha \not \in B$ and $\beta \in C$ does imply that $\alpha \not \in B \times C$, but the converse isn't true since we may have $\beta \not \in C$.

REPLY [1 votes]: Very ugly, but here is:
\begin{align}
(\alpha,\beta) \in [(A \times C) &\setminus (B \times C)] \setminus (A \times D) \\
&\Leftrightarrow \quad (\alpha,\beta) \in (A \times C) \setminus (B \times C) \ \ \& \ \  (\alpha,\beta) \notin A \times D \\
&\Leftrightarrow \quad (\alpha,\beta) \in A \times C \ \ \& \ \ (\alpha,\beta) \notin B \times C \ \ \& \ \ (\alpha,\beta) \notin A \times D \\
&\Leftrightarrow \quad [\alpha \in A \ \ \& \ \ \beta \in C] \ \ \& \ \ [\alpha \notin B \ \ \vee \ \ \beta \notin C] \ \ \& \ \ [\alpha \notin A \ \ \vee \ \ \beta \notin D] \\
&\Leftrightarrow \quad [\alpha \in A \ \ \& \ \ \beta \in C] \ \ \& \ \ [\alpha \notin B] \ \ \& \ \ [\beta \notin D] \\
&\Leftrightarrow \quad (\alpha,\beta) \in (A \setminus B) \times (C \setminus D).
\end{align}
Note : $\&$ stands for $``\textrm{ and}"$ and $\vee$ for $``\textrm{or}"$.


TITLE: What $\oint\vec{B}\cdot d\vec{l}=51.5\mu T\cdot m$ mean?
QUESTION [1 upvotes]: I've been stuck on this practice exam question. I'm supposed to find current through a loop using Ampere's law. But instead of $\mu_0$ it says $\mu$ and it's made me very confused because I can't get the right answer. Any help please?

REPLY [4 votes]: That $\mu$ is the prefix for micro, meaning $10^{-6}$.
The left hand side of that equation is telling you to integrate $\vec{B}\cdot \vec{\operatorname{d}\ell}$ around a closed loop. Usually, you'll have a loop where $|\vec{B}|$ is constant and its direction is along the loop, so you can just multiply $B$ times the perimeter of the loop. Usually.
Can't say more without more details on the problem.


TITLE: Groups of integers with equal sums, equal sums of squares, equal sums of cubes....
QUESTION [4 upvotes]: The following three groups of 6 numbers have equal sums, sums of squares, sums of cubes,... up to fifth powers. How would one go about finding such groups?
784 134
17  901
342 576
12  906
596 322
769 149
684 234
917 1
226 692
Trying to relate this to Newton's identities for sums of roots I have noticed that, when the numbers are taken as roots of polynomials, the polynomials produced are identical up to a constant:
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 316988249001984$
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 239069505576384$
$x^6 - 2754 x^5 + 2845537 x^4 - 1356302772 x^3 + 292772763028 x^2 - 23245284585024 x + 22953865281984$
What is going on here?

REPLY [2 votes]: The coefficients of a monic polynomial can be expressed as functions of the sums of the powers of the roots. And in this case the sums of the powers of the roots are all equal. So the coefficients are equal. (The constant terms are not equal because they involve the sum of the sixth powers of the roots, which are not equal.)
Take as an example the polynomial
$$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$$
The coefficient of $x^2$ is $-(a+b+c)$.
The coefficient of $x$ is $ab+bc+ca=\frac12((a+b+c)^2-(a^2+b^2+c^2))$.
And the constant term is $-abc=-\frac16(a+b+c)^3+\frac12(a+b+c)(a^2+b^2+c^2)-\frac13(a^3+b^3+c^3)$.