A mathematician’s foray into signal processing Carlos Beltr´ an Universidad de Cantabria, Santander, Spain From Complexity to Dynamics: A conference celebrating the work of Mike Shub Carlos Beltr´ an A foray into SP
Credits This work has been greatly inspired by Mike’s thoughts and works Coautors: ´ Oscar Gonz´ alez and Rafael Santamar´ ıa Carlos Beltr´ an A foray into SP
How is it possible? 20 people can use their mobiles at the same time in the same room Carlos Beltr´ an A foray into SP
How is it possible? 20 people can use their mobiles at the same time in the same room Carlos Beltr´ an A foray into SP
How is it possible? 20 people can use their mobiles at the same time in the same room Carlos Beltr´ an A foray into SP
Why 0 , 1 sequences are waves? And one reason for engineers to know complex numbers Carlos Beltr´ an A foray into SP
Why 0 , 1 sequences are waves? And one reason for engineers to know complex numbers Carlos Beltr´ an A foray into SP
Why 0 , 1 sequences are waves? And one reason for engineers to know complex numbers Carlos Beltr´ an A foray into SP
Why 0 , 1 sequences are waves? And one reason for engineers to know complex numbers Carlos Beltr´ an A foray into SP
So you can send a vector in C N , N the number of “antennas” And your friend receives a linear modification of it Carlos Beltr´ an A foray into SP
So you can send a vector in C N , N the number of “antennas” And your friend receives a linear modification of it Carlos Beltr´ an A foray into SP
Interference Alignment: an idea of Jafar’s and Khandani’s research groups Each phone must do some linear algebra Carlos Beltr´ an A foray into SP
Interference Alignment: an idea of Jafar’s and Khandani’s research groups Each phone must do some linear algebra Carlos Beltr´ an A foray into SP
Interference Alignment: an idea of Jafar’s and Khandani’s research groups Each phone must do some linear algebra Carlos Beltr´ an A foray into SP
Interference Alignment: an idea of Jafar’s and Khandani’s research groups Each phone must do some linear algebra Carlos Beltr´ an A foray into SP
The full problem After engineering considerations have been taken into Let K be the number of transmitters/receivers. Let Φ = { ( k , ℓ ) : transmitter ℓ interfers receiver k } ⊆ { 1 , . . . , K } 2 . Let transmitter ℓ have M ℓ antennas, receiver k have N k antennas. Let d j ≤ min { M j , N j } , 1 ≤ j ≤ K , and let H k ℓ ∈ M N k × M ℓ ( C ) be fixed (known). Do there exist U k ∈ M M k × d k ( C ), 1 ≤ k ≤ K and V ℓ ∈ M N ℓ × d ℓ ( C ), 1 ≤ ℓ ≤ K such that U T k H k ℓ V ℓ = 0 ∈ M d k × d ℓ ( C ) , k � = ℓ ? Equivalently, compute the maximal d j that you can use (degrees of Freedom=what SP guys want). This problem has been open since 2006. About 60 research papers. Carlos Beltr´ an A foray into SP
Seen Mike’s Complexity papers? you’ve seen this before So our question is: is π − 1 1 ( H kl ) = ∅ ? For which choices of ( H kl ) ( k , l ) ∈ Φ ? Carlos Beltr´ an A foray into SP
Seen Mike’s Complexity papers? you’ve seen this before So our question is: is π − 1 1 ( H kl ) = ∅ ? For which choices of ( H kl ) ( k , l ) ∈ Φ ? This “double fibration” scheme is a whole business in complex- ity theory and numerical analysis. See for example the works of Shub, Smale and many others by other authors like Armentano, B., Boito, Burgisser, Cucker, Dedieu, Kim, Leykin, Malajovich, Marsten, Pardo, Renegar, Rojas, Shutherland... and others. Carlos Beltr´ an A foray into SP
Compute some dimensions The only non–elementary task follows from the preimage theorem V is a manifold, and � dim C H = ( N k M l − 1) . ( k , l ) ∈ Φ � dim C S = ( d j ( N j + M j − 2 d j )) . 1 ≤ j ≤ K � + � N k d k − d 2 dim C V = N k M l − d k d l k ( k , l ) ∈ Φ k ∈ Φ R � M l d l − d 2 − ♯ (Φ) . + l l ∈ Φ T Carlos Beltr´ an A foray into SP
Are you paying attention? The problem is therefore solved: ◮ If dim H > dim V there is no hope that the problem can be solved for generic ( H kl ) ∈ H . ◮ If dim H ≤ dim V the problem can be solved for generic ( H kl ) ∈ H , because we are in complex and algebraic situations. ... So 60 papers can be summarized with a dimension count argument. The simple case that every transmitter has M , every receiver has N antennas and d degrees of freedom are reached, this dimension count reads: ( K + 1) d ≤ M + N , K the maximum number of users Carlos Beltr´ an A foray into SP
A projection between equal dimensions whose image is a zero measure set But this WON’T happen in real–life problems like the one here, right? Carlos Beltr´ an A foray into SP
Real life problems are indeed singular many times This makes life harder and talks longer Recall that ( K + 1) d ≤ M + N , K the maximum number of users is a necessary condition for generic feasibility of the Interference Alignment. Carlos Beltr´ an A foray into SP
Real life problems are indeed singular many times This makes life harder and talks longer Recall that ( K + 1) d ≤ M + N , K the maximum number of users is a necessary condition for generic feasibility of the Interference Alignment. But... M = N = 3 , d = 2 , K = 2 satisfies this condition and is known NOT to be generically feasible. Carlos Beltr´ an A foray into SP
Real life problems are indeed singular many times This makes life harder and talks longer Recall that ( K + 1) d ≤ M + N , K the maximum number of users is a necessary condition for generic feasibility of the Interference Alignment. But... M = N = 3 , d = 2 , K = 2 satisfies this condition and is known NOT to be generically feasible. So, we have to be more serious. Carlos Beltr´ an A foray into SP
A little gift Carlos Beltr´ an A foray into SP
A little gift Carlos Beltr´ an A foray into SP
A little gift Carlos Beltr´ an A foray into SP
A little gift Carlos Beltr´ an A foray into SP
A general mathematical truth and two examples � � local property → global property. topological constraint Carlos Beltr´ an A foray into SP
A general mathematical truth and two examples � � local property → global property. topological constraint Two examples: � Vector field admits local solution � → solution exists for all t > 0 Manifold is compact Carlos Beltr´ an A foray into SP
A general mathematical truth and two examples � � local property → global property. topological constraint Two examples: � Vector field admits local solution � → solution exists for all t > 0 Manifold is compact � hyperbolicity � → ergodic accesibility Carlos Beltr´ an A foray into SP
One of the most beautiful theorems I have seen For the bored listener: which is yours? Theorem (Ehresmann 1951) Let X , Y be smooth manifolds with Y connected. Let U ⊆ X be a nonempty open subset of X, and let π : U → Y satisfy: Carlos Beltr´ an A foray into SP
One of the most beautiful theorems I have seen For the bored listener: which is yours? Theorem (Ehresmann 1951) Let X , Y be smooth manifolds with Y connected. Let U ⊆ X be a nonempty open subset of X, and let π : U → Y satisfy: ◮ π is a submersion. ◮ π is proper, i.e. π − 1 ( compact ) = compact. Then, π : U → Y is a fiber bundle. In particular, it is surjective. Carlos Beltr´ an A foray into SP
One of the most beautiful theorems I have seen For the bored listener: which is yours? Theorem (Ehresmann 1951) Let X , Y be smooth manifolds with Y connected. Let U ⊆ X be a nonempty open subset of X, and let π : U → Y satisfy: ◮ π is a submersion. ◮ π is proper, i.e. π − 1 ( compact ) = compact. Then, π : U → Y is a fiber bundle. In particular, it is surjective. Corollary If additionally we assume dim( X ) = dim( Y ) then π is a covering map. In particular, the number of preimages of every y ∈ Y is finite and constant. Carlos Beltr´ an A foray into SP
Seen Mike’s Complexity papers? you’ve seen this before So our question is: is π − 1 1 ( H kl ) = ∅ ? For which choices of ( H kl ) ( k , l ) ∈ Φ ? Carlos Beltr´ an A foray into SP
We apply Ehresmann’s theorem to the first projection π 1 to see if it is surjective Let U = { regular points of π 1 . The restriction π 1 | U is a surjection of course. Carlos Beltr´ an A foray into SP
We apply Ehresmann’s theorem to the first projection π 1 to see if it is surjective Let U = { regular points of π 1 . The restriction π 1 | U is a surjection of course. But, it is not proper Carlos Beltr´ an A foray into SP
We apply Ehresmann’s theorem to the first projection π 1 to see if it is surjective Let U = { regular points of π 1 . The restriction π 1 | U is a surjection of course. But, it is not proper Let Σ ⊆ H be the set of singular values of π 1 . Let U = π − 1 1 ( H \ Σ). Again, π 1 | U is surjective. And, this time, it is not difficult to see that it is also proper. Carlos Beltr´ an A foray into SP
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