Symmetric Tensor Decompositions Kristian Ranestad University of - PowerPoint PPT Presentation

Symmetric Tensor Decompositions Kristian Ranestad University of Oslo Linz, 26.11.13 Kristian Ranestad Symmetric Tensor Decompositions

Given F 2 S d = C [ x 0 , . . . , x n ] d homogeneous of degree d . A presentation F = l d 1 + . . . + l d r , with l i 2 S 1 , is called a Waring decomposition of length r of F . Kristian Ranestad Symmetric Tensor Decompositions

Given F 2 S d = C [ x 0 , . . . , x n ] d homogeneous of degree d . A presentation F = l d 1 + . . . + l d r , with l i 2 S 1 , is called a Waring decomposition of length r of F . Question What is the minimal r, the so called rank r ( F ) of F, such that F has a Waring decomposition of length r? Kristian Ranestad Symmetric Tensor Decompositions

Given F 2 S d = C [ x 0 , . . . , x n ] d homogeneous of degree d . A presentation F = l d 1 + . . . + l d r , with l i 2 S 1 , is called a Waring decomposition of length r of F . Question What is the minimal r, the so called rank r ( F ) of F, such that F has a Waring decomposition of length r? How many distinct decompositions of this length does F have? Kristian Ranestad Symmetric Tensor Decompositions

Given F 2 S d = C [ x 0 , . . . , x n ] d homogeneous of degree d . A presentation F = l d 1 + . . . + l d r , with l i 2 S 1 , is called a Waring decomposition of length r of F . Question What is the minimal r, the so called rank r ( F ) of F, such that F has a Waring decomposition of length r? How many distinct decompositions of this length does F have? Can we find r ( F ) , and if not, why? Kristian Ranestad Symmetric Tensor Decompositions

example: conic A smooth conic in CP 2 have equation x 2 0 + x 2 1 + x 2 2 = 0 Kristian Ranestad Symmetric Tensor Decompositions

example: conic A smooth conic in CP 2 have equation x 2 0 + x 2 1 + x 2 2 = 0 The rank is 3, but in how many ways? x 0 x 1 x 2 = 0 defines a polar triangle. Kristian Ranestad Symmetric Tensor Decompositions

Polar triangle F D E Kristian Ranestad Symmetric Tensor Decompositions

Degenerate polar triangle E D Kristian Ranestad Symmetric Tensor Decompositions

Further degenerate polar triangle D Kristian Ranestad Symmetric Tensor Decompositions

Apolarity Sylvester et al. introduced apolarity to find decompositions. Let T = C [ y 0 , . . . , y n ] act on S by di ff erentiation: y i ( F ) = ∂ F . Then ∂ x i X a i x i and g 2 T d ) g ( l d ) = λ g ( a 0 , . . . , a n ) , l = for some λ 6 = 0. Kristian Ranestad Symmetric Tensor Decompositions

Apolarity Sylvester et al. introduced apolarity to find decompositions. Let T = C [ y 0 , . . . , y n ] act on S by di ff erentiation: y i ( F ) = ∂ F . Then ∂ x i X a i x i and g 2 T d ) g ( l d ) = λ g ( a 0 , . . . , a n ) , l = for some λ 6 = 0. Definition g 2 T is apolar to F 2 S if deg g  deg F and g ( F ) = 0. Kristian Ranestad Symmetric Tensor Decompositions

Apolarity Sylvester et al. introduced apolarity to find decompositions. Let T = C [ y 0 , . . . , y n ] act on S by di ff erentiation: y i ( F ) = ∂ F . Then ∂ x i X a i x i and g 2 T d ) g ( l d ) = λ g ( a 0 , . . . , a n ) , l = for some λ 6 = 0. Definition g 2 T is apolar to F 2 S if deg g  deg F and g ( F ) = 0. F ⊥ = { g 2 T | g ( F ) = 0 } ⇢ T . Our key object: The quotient T / F ⊥ is Artinian and Gorenstein . Kristian Ranestad Symmetric Tensor Decompositions

Apolarity lemma Let P ( S 1 ) and P ( T 1 ) denote the projective spaces of 1-dimensional subspaces of S 1 (resp. T 1 ). By apolarity, P ( S 1 ) = P ( T 1 ) ∗ and Γ ⇢ P ( S 1 ) ) I Γ ⇢ T . Kristian Ranestad Symmetric Tensor Decompositions

Apolarity lemma Let P ( S 1 ) and P ( T 1 ) denote the projective spaces of 1-dimensional subspaces of S 1 (resp. T 1 ). By apolarity, P ( S 1 ) = P ( T 1 ) ∗ and Γ ⇢ P ( S 1 ) ) I Γ ⇢ T . Definition Γ ⇢ P ( S 1 ) is an apolar subscheme to F if I Γ ⇢ F ⊥ . Lemma Let Γ = { [ l 1 ] , . . . , [ l r ] } ⇢ P ( S 1 ) , a collection of r points. Then F = λ 1 l d 1 + . . . + λ r l d with λ i 2 C r if and only if I Γ ⇢ F ⊥ ⇢ T . Kristian Ranestad Symmetric Tensor Decompositions

example: conic revisited 2 ) ⊥ = ( y 0 y 1 , y 0 y 2 , y 1 y 2 , y 3 ( x 2 0 + x 2 1 + x 2 0 � y 3 1 , y 3 1 � y 3 2 ) ⇢ T Kristian Ranestad Symmetric Tensor Decompositions

example: conic revisited 2 ) ⊥ = ( y 0 y 1 , y 0 y 2 , y 1 y 2 , y 3 ( x 2 0 + x 2 1 + x 2 0 � y 3 1 , y 3 1 � y 3 2 ) ⇢ T Γ = { [ x 0 ] , [ x 1 ] , [ x 2 ] } ⇢ P ( S 1 ) , I Γ = ( y 0 y 1 , y 0 y 2 , y 1 y 2 ) , so Γ is apolar to x 2 0 + x 2 1 + x 2 2 . Kristian Ranestad Symmetric Tensor Decompositions

Binary forms F 2 C [ x 0 , x 1 ] ) F ⊥ = ( g 1 , g 2 ) ⇢ C [ y 0 , y 1 ] [ Sylvester ] where deg g 1 + deg g 2 = deg F + 2. Kristian Ranestad Symmetric Tensor Decompositions

Binary forms F 2 C [ x 0 , x 1 ] ) F ⊥ = ( g 1 , g 2 ) ⇢ C [ y 0 , y 1 ] [ Sylvester ] where deg g 1 + deg g 2 = deg F + 2. F = λ 1 l d 1 + . . . + λ r l d , I { [ l 1 ] ,..., [ l r ] } = ( g ) ⇢ ( g 1 , g 2 ) . r Kristian Ranestad Symmetric Tensor Decompositions

Binary forms F 2 C [ x 0 , x 1 ] ) F ⊥ = ( g 1 , g 2 ) ⇢ C [ y 0 , y 1 ] [ Sylvester ] where deg g 1 + deg g 2 = deg F + 2. F = λ 1 l d 1 + . . . + λ r l d , I { [ l 1 ] ,..., [ l r ] } = ( g ) ⇢ ( g 1 , g 2 ) . r Assume deg g 1  deg g 2 . ⇢ deg g 1 when g 1 is squarefree r ( F ) = deg g 2 else Kristian Ranestad Symmetric Tensor Decompositions

Cactus rank Definition The cactus rank (or length ) of F is the minimal length of a 0-dimensional apolar subscheme Γ to F , i.e. cr ( F ) := min { length Γ | dim Γ = 0 , I Γ ⇢ F ⊥ } . Kristian Ranestad Symmetric Tensor Decompositions

Cactus rank Definition The cactus rank (or length ) of F is the minimal length of a 0-dimensional apolar subscheme Γ to F , i.e. cr ( F ) := min { length Γ | dim Γ = 0 , I Γ ⇢ F ⊥ } . Clearly cr ( F )  r ( F ) and the inequality may be strict: If d > 2, ) ⊥ = ( y 2 ( x 0 x d − 1 0 , y d 1 ) ) cr ( x 0 x d − 1 ) = 2 < r ( x 0 x d − 1 ) = d . 1 1 1 Kristian Ranestad Symmetric Tensor Decompositions

Border rank Another much studied rank is the border rank : br ( F ) = min { r | F is the limit of forms of rank r } The border rank may be smaller than the cactus rank. Kristian Ranestad Symmetric Tensor Decompositions

Bounds on the rank The most famous bound on the rank is not a bound Theorem (Alexander-Hirschowitz 1995) Let F 2 C [ x 0 , . . . , x n ] be a general form of degree d, then 1 � n + d � r(F)= AH(d,n):= d e , except n +1 n r(F)=n+1 if d=2, r(F)=6,10,15 if d=4, n=2,3,4, r(F)=8 if d=3, n=4. Remark Special forms may have larger rank, a sharp upper bound is only known in a few special cases (( n , d ) = ( n , 2) , (1 , d ) , (2 , 3) , (2 , 4) , (3 , 3)) . Kristian Ranestad Symmetric Tensor Decompositions

The simplest lower bound for the rank is explained by di ff erentiation . If F = l d 1 + . . . + l d and g 2 C [ y 0 , . . . , y n ] s r then g ( F ) = λ 1 l d − s + . . . + λ r l d − s for some λ 1 , . . . , λ r 2 C 1 r Kristian Ranestad Symmetric Tensor Decompositions

The simplest lower bound for the rank is explained by di ff erentiation . If F = l d 1 + . . . + l d and g 2 C [ y 0 , . . . , y n ] s r then g ( F ) = λ 1 l d − s + . . . + λ r l d − s for some λ 1 , . . . , λ r 2 C 1 r So, the dimension h ( F , s ) of the vector space of partials of order s is a lower bound for r ( F ). In fact r ( F ) � max { h ( F , s ) | 0 < s < d } The same lower bound is valid for the cactus rank and the border rank. Kristian Ranestad Symmetric Tensor Decompositions

An improvement by Landsberg and Teitler depending on the singular locus of the hypersurface V ( F ). Let d ( F , s ) be the dimension of the locus of points on V ( F ) of multiplicity at least s . Kristian Ranestad Symmetric Tensor Decompositions

An improvement by Landsberg and Teitler depending on the singular locus of the hypersurface V ( F ). Let d ( F , s ) be the dimension of the locus of points on V ( F ) of multiplicity at least s . Theorem (Landsberg-Teitler 2009) Let F 2 C [ x 0 , . . . , x n ] d . Assume that V ( F ) is not a cone, and let 0 < s < d. Then r ( F ) � h ( F , s ) + d ( F , s ) + 1 . The proof uses apolarity in a very essential way. Kristian Ranestad Symmetric Tensor Decompositions

Bounds on the cactus rank For n > 2 and d > 6 the cactus rank of a general form is smaller than the rank: Proposition (Bernardi,R 2011) Let F 2 C [ x 0 , . . . , x n ] d be any form of degree d, then � n + k ⇢ � 2 when d = 2 k + 1 cr ( F )  N ( d , n ) := n � n + k � n + k +1 � � + when d = 2 k + 2 n n Notice that N ( d , n ) ⇡ O (( d / 2) n ), while AH ( d , n ) ⇡ O ( d n ). Question Is this bound sharp? (Yes, at least for cubics (nov 13)) Kristian Ranestad Symmetric Tensor Decompositions

The proof of the Proposition uses: Theorem (Emsalem 1978) Let Γ ⇢ C n be a local 0 -dimensional scheme with I Γ ⇢ ( y 1 , . . . , y n ) . ) 9 f 2 C [ x 1 , . . . , x n ] s . t . I Γ = f ⊥ . Γ is Gorenstein ( Furthermore, in this case, length Γ = dim C D ( f ) where D ( f ) ⇢ C [ x 1 , . . . , x n ] is the space of partial derivatives of f of all orders. Kristian Ranestad Symmetric Tensor Decompositions