Sub Banach spaces (Banach algebras) of the disc algebra which are invariant under the differentiation operator

Is there a complete classification of all Banach subspace of the disc algebra $ \mathcal{A}(\mathbb{D})$ which are invariants under the differentiation operator? Is the a complete classification of such Banach subspace for which the differentiation is a bounded operator/

Is there a complete classification of all Banach sub algebra of $ \mathcal{A}(\mathbb{D})$ which are invariant under the differentiation operator?

Having these questions in my mind, I arrived at this question

Is there a holomorphic function on open unit disc with this property?

Cartan’s magic formula for diffferential graded algebra


Algebra $ A$ is called graded algebra if it has a direct sum decomposition $ A=\bigoplus_{k\in\Bbb Z} A^k$ s.t. product satisfies $ (A^k)(A^l)\subseteq(A^{k+l}) \text{ for each } k, l.$

A differential graded algebra is graded algebra with chain complex structure $ d \circ d = 0$ .

Derivation of degree $ k$ on $ A$ means a linear map $ D:A \to A$ s.t. $ $ D(A_j)\subset A_{j+k} \text{ and } D(ab)=(Da)b + (-1)^{ik}a(Db), a\in A_i$ $

All smooth forms on $ n$ -manifold $ M$ is a differential graded algebra $ \Omega^{\bullet}(M)=\bigoplus_{k=0}^{n} \Omega^k(M)$ , with wedge product and exterior derivative.

In proving Cartan’s magic formula $ \mathcal{L}_X=i_X \circ d + d\circ i_X$ holds for $ \Omega^{\bullet}(M)$ , we can use the following steps:

  1. Show that two degree $ 0$ derivations on $ \Omega^{\bullet}(M)$ commuting with $ d$ are equal iff they agree on $ \Omega^0(M)$ .

  2. Show that $ \mathcal{L}_X$ and $ i_X \circ d + d \circ i_X$ are derivations on $ \Omega^{\bullet}(M)$ commuting with $ d$ .

  3. Show that $ \mathcal{L}_X f = Xf = i_Xdf+ d i_Xf$ for all $ f \in C^{\infty}(M)=\Omega^0(M)$ .

My question:

  1. Why step 1 is ture? Why commuting with $ d$ is so important?

  2. Can step 1 be extended to any derivations without restriction on degree.

Thank you.

Relational Algebra with only one operator?

There’s a parlour game of inventing exotic operators for Relational Algebra, and thereby reducing the number of operators needed to be ‘Relationally Complete’. A popular operator for this is ‘Inner Union’ aka SQL’s UNION CORRESPONDING.

I’ve just bumped into a single-operator basis for FOL, due to Schönfinkel. It’s a combo of Sheffer stroke (written infix |) and existential quant (with the bound var superscripted).

P(x) |x Q(x) ≡ ¬∃x.(P(x)∧Q(x))

Q 1. Could there be a Relational Operator corresponding to that?

Q 2. If so, does that mean there could be a version of Relational Algebra with only one operator?

Q 3. If not, in what sense is Codd’s 1972 set “complete”?

My thoughts so far:

Q 1. No. The FOL corresponds OK to RA (Natural Join). The corresponds OK to ‘Remove’ aka project-away, sometimes written π-hat. But RA can only express correspondence to negation when ¬ is nested inside . I.e. FOL P(x) ∧ ¬Q(x) corresponds to RA P MINUS Q. Whereas this single FOL operator has ¬ at outer level (i.e. absolute complement, not relative).

The reason Codd doesn’t allow absolute complement is it makes queries ‘unsafe’, that is domain-dependent.

Q 2. Then no. Supplementary q: it’s well known Codd omitted RENAME/ρ from his original set. Rename is needed to translate a FOL expression using = between variables:

∃x. P(x) ∧ (x = y)       -- corresponds to  ρ{y := x }(P)            -- relation P with attrib x 

Presumably Schönfinkel’s operator doesn’t avoid the need for =(?).

Q 3. Then how does Codd’s original RA express an equivalent to a FOL expression with outermost ¬? Or outermost , which is the same thing:

∀y.Q(y)≡¬∃y.¬Q(y) 

Von Neumann algebra such that every state is normal- what can be said about its dimension? What about the other direction?

This is exercise 5 in page 130 of Masamichi Takesaki ‘s Theory of Operator Algebras 1 (TO SHOW THAT IT IS FINITE DIMENSIONAL, AND THAT THE IFF HOLDS).

I know that now every state is weakly continuous on the unit ball. Maybe that means that it must be separable, and then I can show that it is finite dimensional?

Thanks in advance!

Abstract Algebra – Showing something to be a group

I’ve been posed the question:

“Let $ P$ be the pairs (a,b) where $ a ∈ Z$ $ 4$ , and $ b ∈ Z$ $ 2$ .

An operation, $ *$ , is defined by: $ (a,b)*(c,d)=(a+c$ (mod 4)$ , b+d $ (mod 2)) for all $ (a,c),(b,d)∈P$

How do I show that this is a group?

I know how to do this with multiplication tables by working through the axioms but I don’t know how to apply these to this question, nor if that’s the best approach