nt.number theory – Is there an analogue for Ramanujan–Serre derivative for Hilbert modular forms?

If $f$ is a modular form of weight $k$, it is well known that
$$
D(f)=f’ -tfrac k{12}E_2f
$$

is modular of weight $k+2$.
Here $E_2$ is the Eisenstein series. I wanted to ask if there is an extension of this fact for Hilbert modular forms.

When looking up Hilbert modular forms (Shimura 1975), I was only able to find the following differential operators: for a Hilbert modular form $f$ on $mathbb H^n$, we define
$$
D_{j,t}(f)=left(frac{t}{2i y_j}+frac{partial}{partial z_j}right)f
$$

where $z_j=x_j+iy_j$ is the $j$-th coordinate. I don’t think this is the same operator (is it?)

I am really looking for a many variable version of the operator $D(f)=f’ -frac k{12}E_2f$. I don’t really have a background in analysis or number theory so this question may be really easy for experts.

euclidean geometry – Deduction of natural numbers from Hilbert axioms

I state that I am not an expert in formal systems, but I did something as a “self-taught” because I am passionate about it. My question is the following: since geometry hilbert axioms are equivalent to the real numbers axioms from the primes one must deduce the natural numbers, the induction theorem, etc. How you do it? If the following is not wrong we must define segment multiples in the same way that we define natural numbers from real numbers axioms. Premise: as did hilbert in his axioms, since the ancient Greeks used the language of logic (and not that of sets) we use the second order logic language. In this way the first step to introduce the multiples of AB is to define the analogue of the notion of inductive set, i.e. the inductive properties (with respect to the segment AB):

Definition 1 A segments property P is inductive with respect to segment $AB$ when:

  1. P$(AB)$
  2. $DE cong AB$ $wedge$ Bet$(C,D,E)to$ (P$(CD)to$ P$(CE)$)

where Bet is the betwennes relation.

Exemple 1 The relation $…geq AB$ is inductive with respect to $AB$.

Proof. in fact $AB geq AB$ so that the 1 of definition 1 is verified. In addition let C, D, E such that $DE cong AB$ $wedge$ Bet$(C,D,E)$. If $CD geq AB$ then $CE geq AB$. Thus also 2 of definition 1 is verified. QED

we are now ready for:

Definition 2 ($AB$ multiple) A segment $RS$ is a multiple of $AB$ if for every inductive property P we have P$(RS)$.

The definition given above is the analogue of the definition of natural numbers as the intersection of every inductive sets of R.

Rem. 1 $AB$ is a multiple of itself. In effect for 1 of definition 1 for every P inductive we have P$(AB)$. For definition 2 we have the assert. QED.

Rem.2 The property M(to be a $AB$ multiple) is inductive.

Proof. For Rem.1 we have M$(AB)$. Let $DEcong AB$ and Bet$(C,D,E)$. If M$(CD)$ for every P inductive we have P$(CD)$. Furthermore, since $DEcong AB$, one must have also P$(CE)$ for 2 of definition 1, in other words M$(CE)$ QED.

Conseguences of of above facts are:

1 $AB$ is multiple of $AB$

2 If $DEcong AB$ , Bet$(C,D,E)$ and M$(CD)$ then M$(CE)$

3 If M$(CE)$ and Bet$(C,D,E)$ and $DEcong AB$ then $CE ncong AB$

4 If M$(CD)$ and M$(PQ)$ and Bet$(C,D,E)$ and Bet$(P,Q,R)$ and $DEcong QR cong AB$ then $CD cong PQ$

5 (Induction Theorem) Let P a property. If there exist a segment $PQ$ such that M$(PQ)$ and P$(PQ)$ and if:

  • P$(AB)$
  • $DEcong AB$
  • Bet$(C,D,E)$
  • P$(CD)to$ P$(CE)$

then P$(HK)$ for every $HK$ such that M$(HK)$.

The problem is that in the geometric proofs, for example Euclid Segement division when the two segment are supposed commensurables , the division of a polygon in n triangles etc.the recursion theorem is not the above (in terms of segment multiples) but it is in terms of integers. Furthermore in the geometric definitions (polygon for example) the induction definition is not the above (in terms of segments multiples) but is in terms of integers. it is therefore necessary “to translate the integers in terms of multiples of a segment”. how to do this?

Identification of products in the tensor product of hilbert spaces

I’m relatively new to tensor products and particularly interested in tensor products of Hilbert spaces. I did read a short note on the explicit construction of theses spaces, which is also covered on Wikipedia (link below). However, I have some general question and hope someone can bring some light into my confusion.

Following the wikipedia article one can construct the tensor product of Hilbert spaces $H_1$ and $H_2$ as the space which is isometrically and linearly isomorphic to $HS(H_{1}^{*},H_{2})$, the space of Hilber-Schmidt operators from $H_1^*$ to $H_2$.

The idea is to identify to every tensor $x_1otimes x_2$ with $x_iin H_i$ and $x^*in H_1^*$ the map

$$x^*mapsto x^*(x_1)x_2$$

On the other hand one often defines the tensor product via a new inner product. For this the tensor product (in algebraic way) is first constructed. Then an inner product is defined. I quote Wikipedia:

Construct the tensor product of $H_1$ and $H_2$ as vector spaces as explained in the article on tensor products. We can turn this vector space tensor product into an inner product space by defining
$$
{displaystyle langle phi _{1}otimes phi _{2},psi _{1}otimes psi _{2}rangle =langle phi _{1},psi _{1}rangle _{1},langle phi _{2},psi _{2}rangle _{2}quad {mbox{for all }}phi _{1},psi _{1}in H_{1}{mbox{ and }}phi _{2},psi _{2}in H_{2}}$$

and extending by linearity. That this inner product is the natural one is justified by the identification of scalar-valued bilinear maps on $H_1 times H_2$ and linear functionals on their vector space tensor product. Finally, take the completion under this inner product. The resulting Hilbert space is the tensor product of $H_1$ and $H_2$.

question 1 In the last definition, if I read it correctly, we build first the tensor product in a algebraic way. Then an inner product is on that tensor product defined. Finally the completion is taken and one sees that this corresponds with the original (algebraic) tensor product? Or how does the algebraic and the one stemming from the completion related?

question 2 I often see that for Hilbert spaces, say of functions, tensor products are defined via multiplication. That is for functions $f_iin H_i$ we have $f_1otimes f_2:=f_1(x)f_2(x)$. How do we get this concrete example of “multiplication”. I’m particular interested in Hilbert spaces of scalar valued functions, if that matters.

question 3. What is the relation of the explicit construction via Hilber-Schmidt operators to the multiplicative case mentioned in question 2.

mp.mathematical physics – Riemannian metric induced by $n$ point sets in Hilbert spaces?

Excuse my naive question:
In quantum mechanics the mathematical description is through Hilbert spaces.
Suppose we have $n$ points $x_1,cdots,x_n$ in this Hilbert-Space.
Then we get the Gramian matrix:

$$G = (<x_i,x_j>)$$

which is symmetric, positive definite and lives in a Riemannian Manifold of Symmetric positive definite matrices.

So each set of $n$ points in this Hilbert Space might be said to induce a point $G$ in the Riemannian manifold.

Is there any physical “reality” for this thought experiment?

Also asked here: https://physics.stackexchange.com/questions/648895/riemannian-metric-induced-by-n-point-sets-in-hilbert-spaces since I was not sure which site might be the better fit.

functional analysis – What is the Hilbert transform of 1?

I am studying singular integrals and I have understood that if we have an operator $T$ which is bounded in $L^2$, then we can extend this operator for functions in $L^{infty}$ to BMO space. This is the case of the Hilbert transform.

$$Hf(x)=frac{1}{pi}text{p.v}int_{mathbb{R}}frac{f(y)}{x-y}dy$$

My question is, what is $H1?$ It is clear that the function $f(x)=1$ is bounded, so $H1$ would be a BMO function, but which function?

Maybe the key is to work with distributions instead of functions, or using the Fourier definition of the Hilbert tranform:

$$widehat{(Hf)}(xi)=-i text{sgn}(xi) widehat{f}(xi)$$

This is:

$$widehat{(H1)}(xi)=-i text{sgn}(xi) delta(xi)$$

but we know its inverse Fourier transform?

Thank you very much

functional analysis – For a compact operator $T in K(X, H)$ and H an Hilbert space, $overline{T(X)}$ is separable

Let X be a normed space, H be a Hilbert space, and let $T in K(X, H)$ the set of compact operators. Show that T(X) is separable. I tried to use the fact that the set of finite rank operators $F(X,H) = K(X,H)$ when H is an Hilbert space, to show that $T(x)$ is countable (since we need to show that $overline{T(X)}$ is separable, that is, that it contains a countable dense subset)

real analysis – Two theorems of existence of a minimizer in a Hilbert space

I am looking for some results of existence of minimizers in infinite dimensional Hilbert spaces. I found (surfing the web) these two results:

${bf Theorem 1.}$ Let $H$ be an Hilbert space and $Csubset H$ be a closed and convex subset. Let $J:Htomathbb{R}$ be a weakly lower semicontinuous functional. If $C$ is (also) bounded or $J$ is coercive, thus exists $min_{uin C} J(u)$.

${bf Theorem 2.}$ Let $H$ be an Hilbert space and $Csubset H$ be a closed and convex subset. Let $J:Ctomathbb{R}$ be a lower semicontinuous convex functional. Thus $J$ is bounded from below and the minimum $min_{uin C}J(u)$ is attained.

I guess the Theorems states (almost) the same thing since:

$J$ lower semicontinuous and convex $implies$ $J$ weakly lower semicontinuous.

The two differences I see are:

  1. In the first case $J:Htomathbb{R}$, while $J:Ctomathbb{R}$ in the second one, could anyone please explain me why?

  2. In the second case, we are able to state that $J$ is bounded from below, what about the first case instead?

My question is: could someone please give me a detailed reference about these result? (better if it contains a proof).

Furthermore, there exists some conditions which ensure the convexness of a functional in a Hilbert space? I am interested in some characterizations.

I hope someone could help. Thank you in advance!

functional analysis – Let the complex hilbert space be $H=mathcal{L}^{2}[0,frac{pi}{2}]$. Consider application T and prove the following.

begin{equation}
forall tin left(0,frac{pi}{2} right) qquad qquad Tf(t)=cos (t)int_{0}^{t}sin(s)f(s)ds
end{equation}

  1. Show that $ Tf $ is a continuous function with $ f in H $. Calculate the $ Tf (t) $ values ​​for $ t = 0 $ and $ t = frac{pi}{2} $.

  2. Show that $ T $ is continuous.

  3. Determine $T^*$. Show that $ T $ and $ T^* $ are compact.

  4. Show that if $ f $ is continuous then $ Tf $ and $ T^*f $ are of class $ C^1 $. Calculate the derivatives of $Tf$ and $T^*f$.

  5. Let’s consider $ A = T + T^* $. Show that $ A $ is compact and self-adjoint.

  6. Show that if $ f in H $ and $ Af = lambda f $, with $ lambda $ not null, then $ f $ is of class $ C ^ {2} $ over $ (0, frac {pi } {2}) $ and verify a second-order linear differential equation of the form $ f ” + alpha (lambda) f = 0 $, with $ alpha (lambda) in mathbb {R} $ . Determine $ f (0) $ and $ f (frac{pi} {2})$.

  7. Deduct the eigenvalues ​​of $ A $. Calculate the norm of $ A $.

I made the literal 1), it assumed the existence of a convergent sequence $ (t_ {n}) in (0, frac { pi} {2}) $ and with thanks to the continuity of the cos and the integral it is It follows that $ lim_{n rightarrow infty} Tf (x_ {n}) = Tf (x) $ so Tf is continuous. And evaluate Tf (t) at each point I got zero.

For the literal 2), I used the fact that T is linear, and to show that it is continuous, it sufficed to show that it is bounded like this
begin{equation}
|Tf(t) | = | cos (t) int_{0}^{t}sin(s)f(s)ds | leq |cos (t)| int_{0}^{t} | sin(s) | |f(s)|dsleq int^{t}_{0}|f(s)|dsleq |1|_{L^{2}}|f|_{L^{2}}=sqrt{frac{pi}{2}}||f||_{L^{2}}
end{equation}

por tanto $T$ es continuo.

Now I have a problem from literal 3) to find $ T^{*} $.