The epic battle between Markov and phylogenetic invariants: equations Jeremy Sumner School of Physical Sciences University of Tasmania, Australia Phylomania 2014 Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 1 / 16
The people I did this with Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 2 / 16
b b b b b b Symmetry in phylogenetics: Sequence and leaf permutations ∆ 1 = { residual for T 1 = A C D } ALGEBRA ALERT B ∆ 2 = { residual for T 2 = A D } B i.e. a “representation” C ∆ 3 = { residual for T 3 = A B C } of S 4 on { T 1 , T 2 , T 3 } D seqA PHYLOGENETIC seqB − → − → (∆ 1 , ∆ 2 , ∆ 3 ) seqC METHOD seqD seqD PHYLOGENETIC seqC − → − → (∆ 1 , ∆ 2 , ∆ 3 ) seqB METHOD seqA seqA PHYLOGENETIC seqD − → − → (∆ 3 , ∆ 1 , ∆ 2 ) seqB METHOD seqC Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 3 / 16
The leaf action ALGEBRA ALERT i.e. “action” of × 4 GL ( n ) with Markov matrices M i seqA seqB − → METHOD − → (∆ 1 , ∆ 2 , ∆ 3 ) − → seqC 1 2 3 4 seqD seqA seqB METHOD − → − → (∆ 1 , ∆ 2 , ∆ 3 )??? − → seqC seqD 1 2 3 4 Ideally tree support should depend only on “internal” part of tree Isn’t this what “phylogenetic invariants” achieve? Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 4 / 16
b b b b b b Phylogenetic invariants A C A B A B T 1 = T 2 = T 3 = B D C D D C Consider polynomials f ( P ) = f ( p AAAA , p AAAC , p AAAG , . . . , p TTTT ) Phylogenetic “invariant”: (Cavender, Felsenstein, Lake, etc.) f ( P 1 ) = 0 f ( P 2 ) � = 0 f ( P 3 ) � = 0 Algebraic statistics (Sturmfels, Pacter, et. al.) : Ideals, varieties, etc. Our perspective: Groups, modules, etc. In either case f becomes an infinite space � f 1 , f 2 , f 3 , . . . � Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 5 / 16
b b b b b b b b Back to sequence and leaf permutations seqA A C A B A B INVARIANTS seqB − → − → D or D or seqC METHOD B C D C seqD Permute seqA ↔ seqB = ⇒ (∆ 1 , ∆ 2 , ∆ 3 ) → (∆ 1 , ∆ 3 , ∆ 2 )? ‘ Biologically symmetric ’ invariants (E 2009, R&H 2012) ‘ Invariant ’ invariants! (F-S pers. comm.) Quartet Stabilizer A C G = S 2 ≀ S 2 B D = � ( AB ) , ( CD ) , ( AC )( BD ) � ALGEBRA ALERT Irreducible representations of G provide distinguished basis for invariants (S&J 2009) Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 6 / 16
From an algebraic point of view, this is only half the story. What about the leaf action? Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 7 / 16
The problem with phylogenetic invariants � f 1 , f 2 , f 3 , . . . � ALGEBRA ALERT Leaf action i.e. “action” of × 4 GL ( n ) − → 1 2 3 4 with Markov matrices M i 1 2 3 4 BIG INSIGHT ijkl = � Linear combination of p ijkl , p ′ coeffs from M i ⇓ i = � Linear combination of f i , f ′ coeffs from M i Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 8 / 16
b b The problem with using phylogenetic invariants � f 1 , f 2 , f 3 , . . . � Tree “residual” ∆ := � i | f i ( P ) | 2 depends on choice of basis { f 1 , f 2 , f 3 . . . } Leaf action − → 1 2 3 4 1 2 3 4 ∆= f 2 1 + f 2 2 + f 2 3 ∆ ′ = f ′ 2 + f ′ 2 + f ′ 2 1 2 3 Any measure ∆ entails a choice of phylogenetic “invariants” equivalent to alternative choice evaluated at a displaced P. i.e. ∆ is not invariant to leaf action Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 9 / 16
Markov invariants solve this problem! Leaf action − → 1 2 3 4 1 2 3 4 i = � Linear combination of f i , THE BIG INSIGHT : f ′ coeffs from the M i Markov invariants: q → λ q Existence theorem (S,C,J,& J, 2009) : λ = products of det( M i ) Surely Markov invariants are good because they don’t depend on “internal” part of the tree? Yes! Log-det and Hadamard q -coordinates; the magical squangles (H,S,& J 2012) . Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 10 / 16
Binary quartet: The simplest case possible 12 | 34 “flattening” of quartet probability distribution P : � p 00 , 00 p 00 , 01 p 00 , 10 p 00 , 11 � p 01 , 00 p 01 , 01 p 01 , 10 p 01 , 11 P = p 10 , 00 p 10 , 01 p 10 , 10 p 10 , 11 p 11 , 00 p 11 , 01 p 11 , 10 p 11 , 11 Initial value on “stubby” T 1 , T 2 , and T 3 : 00 01 10 11 00 01 10 11 00 ∗ 0 0 ∗ 00 ∗ 0 0 0 01 0 0 0 0 01 0 ∗ 0 0 P 1 = P 2 = P 3 = 10 0 0 0 0 10 0 0 ∗ 0 11 ∗ 0 0 ∗ 11 0 0 0 ∗ Notice 3 × 3 minors are 0 on T 1 and non-zero on T 2 and T 3 LEAF ACTION+BIG INSIGHT = ⇒ minors are phylo invariants Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 11 / 16
b b b b b b b b Minors and leaf permutations DETAIL OPTIONAL B C Rotation: A C − → B D A D C A A C Reflection: − → B D D B � 1 0 0 0 � → KP and P t , where K = 0 0 1 0 Flattening: P − 0 1 0 0 0 0 0 1 3 × 3 minors under leaf permutations: 00 01 10 11 00 ◦ ♦ ♦ � 01 ♦ ⋆ ⋆ � 10 ♦ ⋆ ⋆ � 11 ∗ � � � Gives six possible leaf perm invariant residuals: ∆ = ◦ 2 + ♦ 2 + � 2 + ⋆ 2 + � 2 + ∗ 2 Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 12 / 16
But what about the leaf action? Leaf action for flattening: P → XPY t (where X = M 1 ⊗ M 2 and Y = M 3 ⊗ M 4 ) Under the leaf action the minors become all mixed up! Leaf action Linear combination of ◦ , ♦ , ◦ , ♦ , � , ⋆ , � , ∗ − → � , ⋆ , � , ∗ , coeffs from the M i BADNESS! Leaf action Markov invariants, “the squangle”: q − → λ q ∆ = q 2 provides a tree-topology residual that is invariant to changes of parameter values at leaves of tree AWESOMENESS! Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 13 / 16
But wait, there’s more: Signed least squares Three flattenings = ⇒ three squangles: q 1 , q 2 , q 3 . Rep theory: q 1 + q 2 + q 3 = 0 use q 3 = − q 2 in place of q 1 = 0. Leaf action gives semi-algebraic constraints: u , v , w > 0 Hypothesis E [ q 1 ] E [ q 2 ] E [ q 3 ] T 1 0 − u u T 2 v 0 − v T 3 − w w 0 u = 1 Least squares estimate: � 2 ( q 3 − q 2 ) or � u = 0. Residuals: ∆ = 1 2 q 2 1 or q 2 2 + q 2 3 Second case sends q 2 , q 3 → 0 as best estimate. Analogous situation for minors Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 14 / 16
Origin of signed least squares Leaf action: P → XPY t (where X = M 1 ⊗ M 2 and Y = M 3 ⊗ M 4 ) On matrix of minors, inverses get in the act: � 1 − a � � + − � ⇒ M - 1 = b M = = 1 − b − + a Leaf action on minors: � ◦ ♦ ♦ � � � + − − + � � ∗ 0 0 0 � � + − − + � ♦ ⋆ ⋆ � − + + − − + + − 0 ∗ 0 0 = − + + − − + + − 0 0 ∗ 0 ♦ ⋆ ⋆ � + − − + 0 0 0 ∗ + − − + � � � ∗ � + − − + � − + + − = − + + − + − − + Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 15 / 16
Take home message ◮ Theory says “Markov invariants ” > “phylogenetic invariants” ◮ In particular, the squangles should give stable tree residual function in face of changing rate parameters. ◮ Over to you Barbara... REFS Eriksson N. 2009. Using invariants for phylogenetic tree construction. IMA Vol. Math. Appl. Rusinko JP, Hipp, B. 2013. Invariant based quartet puzzling. Alogrithms for Molecular Biology Sumner JG, Charleston MA, Jermiin LS, Jarvis PD. 2008. Markov invariants, plethysms, and phylogenetics. JTB Sumner JG, Jarvis PD. 2009. Markov invariants and the isotropy group of a quartet. JTB Holland BR, Sumner JG, Jarvis PD. 2013. Low-Parameter Phylogenetic Inference Under the GM Model. Syst. Biol. Jeremy Sumner The epic battle between Markov and phylogenetic invariants: equations 16 / 16
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