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Rename _rho to _restrict_to_space and make it work on subspaces instead.
[sage.d.git] / mjo / cone / cone.py
1 # Sage doesn't load ~/.sage/init.sage during testing (sage -t), so we
2 # have to explicitly mangle our sitedir here so that "mjo.cone"
3 # resolves.
4 from os.path import abspath
5 from site import addsitedir
6 addsitedir(abspath('../../'))
7
8 from sage.all import *
9
10
11 def _basically_the_same(K1, K2):
12 r"""
13 Test whether or not ``K1`` and ``K2`` are "basically the same."
14
15 This is a hack to get around the fact that it's difficult to tell
16 when two cones are linearly isomorphic. We have a proposition that
17 equates two cones, but represented over `\mathbb{Q}`, they are
18 merely linearly isomorphic (not equal). So rather than test for
19 equality, we test a list of properties that should be preserved
20 under an invertible linear transformation.
21
22 OUTPUT:
23
24 ``True`` if ``K1`` and ``K2`` are basically the same, and ``False``
25 otherwise.
26
27 EXAMPLES:
28
29 Any proper cone with three generators in `\mathbb{R}^{3}` is
30 basically the same as the nonnegative orthant::
31
32 sage: K1 = Cone([(1,0,0), (0,1,0), (0,0,1)])
33 sage: K2 = Cone([(1,2,3), (3, 18, 4), (66, 51, 0)])
34 sage: _basically_the_same(K1, K2)
35 True
36
37 Negating a cone gives you another cone that is basically the same::
38
39 sage: K = Cone([(0,2,-5), (-6, 2, 4), (0, 51, 0)])
40 sage: _basically_the_same(K, -K)
41 True
42
43 TESTS:
44
45 Any cone is basically the same as itself::
46
47 sage: K = random_cone(max_ambient_dim = 8)
48 sage: _basically_the_same(K, K)
49 True
50
51 After applying an invertible matrix to the rows of a cone, the
52 result should be basically the same as the cone we started with::
53
54 sage: K1 = random_cone(max_ambient_dim = 8)
55 sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular')
56 sage: K2 = Cone( [ A*r for r in K1.rays() ], lattice=K1.lattice())
57 sage: _basically_the_same(K1, K2)
58 True
59
60 """
61 if K1.lattice_dim() != K2.lattice_dim():
62 return False
63
64 if K1.nrays() != K2.nrays():
65 return False
66
67 if K1.dim() != K2.dim():
68 return False
69
70 if K1.lineality() != K2.lineality():
71 return False
72
73 if K1.is_solid() != K2.is_solid():
74 return False
75
76 if K1.is_strictly_convex() != K2.is_strictly_convex():
77 return False
78
79 if len(LL(K1)) != len(LL(K2)):
80 return False
81
82 C_of_K1 = discrete_complementarity_set(K1)
83 C_of_K2 = discrete_complementarity_set(K2)
84 if len(C_of_K1) != len(C_of_K2):
85 return False
86
87 if len(K1.facets()) != len(K2.facets()):
88 return False
89
90 return True
91
92
93
94 def _restrict_to_space(K, W):
95 r"""
96 Restrict this cone a subspace of its ambient space.
97
98 INPUT:
99
100 - ``W`` -- The subspace into which this cone will be restricted.
101
102 OUTPUT:
103
104 A new cone in a sublattice corresponding to ``W``.
105
106 EXAMPLES:
107
108 When this cone is solid, restricting it into its own span should do
109 nothing::
110
111 sage: K = Cone([(1,)])
112 sage: _restrict_to_space(K, K.span()) == K
113 True
114
115 A single ray restricted into its own span gives the same output
116 regardless of the ambient space::
117
118 sage: K2 = Cone([(1,0)])
119 sage: K2_S = _restrict_to_space(K2, K2.span()).rays()
120 sage: K2_S
121 N(1)
122 in 1-d lattice N
123 sage: K3 = Cone([(1,0,0)])
124 sage: K3_S = _restrict_to_space(K3, K3.span()).rays()
125 sage: K3_S
126 N(1)
127 in 1-d lattice N
128 sage: K2_S == K3_S
129 True
130
131 TESTS:
132
133 The projected cone should always be solid::
134
135 sage: set_random_seed()
136 sage: K = random_cone(max_ambient_dim = 8)
137 sage: _restrict_to_space(K, K.span()).is_solid()
138 True
139
140 And the resulting cone should live in a space having the same
141 dimension as the space we restricted it to::
142
143 sage: set_random_seed()
144 sage: K = random_cone(max_ambient_dim = 8)
145 sage: K_P = _restrict_to_space(K, K.dual().span())
146 sage: K_P.lattice_dim() == K.dual().dim()
147 True
148
149 This function should not affect the dimension of a cone::
150
151 sage: set_random_seed()
152 sage: K = random_cone(max_ambient_dim = 8)
153 sage: K.dim() == _restrict_to_space(K,K.span()).dim()
154 True
155
156 Nor should it affect the lineality of a cone::
157
158 sage: set_random_seed()
159 sage: K = random_cone(max_ambient_dim = 8)
160 sage: K.lineality() == _restrict_to_space(K, K.span()).lineality()
161 True
162
163 No matter which space we restrict to, the lineality should not
164 increase::
165
166 sage: set_random_seed()
167 sage: K = random_cone(max_ambient_dim = 8)
168 sage: S = K.span(); P = K.dual().span()
169 sage: K.lineality() >= _restrict_to_space(K,S).lineality()
170 True
171 sage: K.lineality() >= _restrict_to_space(K,P).lineality()
172 True
173
174 If we do this according to our paper, then the result is proper::
175
176 sage: set_random_seed()
177 sage: K = random_cone(max_ambient_dim = 8)
178 sage: K_S = _restrict_to_space(K, K.span())
179 sage: K_SP = _restrict_to_space(K_S.dual(), K_S.dual().span()).dual()
180 sage: K_SP.is_proper()
181 True
182 sage: K_SP = _restrict_to_space(K_S, K_S.dual().span())
183 sage: K_SP.is_proper()
184 True
185
186 Test the proposition in our paper concerning the duals and
187 restrictions. Generate a random cone, then create a subcone of
188 it. The operation of dual-taking should then commute with
189 _restrict_to_space::
190
191 sage: set_random_seed()
192 sage: J = random_cone(max_ambient_dim = 8)
193 sage: K = Cone(random_sublist(J.rays(), 0.5), lattice=J.lattice())
194 sage: K_W_star = _restrict_to_space(K, J.span()).dual()
195 sage: K_star_W = _restrict_to_space(K.dual(), J.span())
196 sage: _basically_the_same(K_W_star, K_star_W)
197 True
198
199 """
200 # First we want to intersect ``K`` with ``W``. The easiest way to
201 # do this is via cone intersection, so we turn the subspace ``W``
202 # into a cone.
203 W_cone = Cone(W.basis() + [-b for b in W.basis()], lattice=K.lattice())
204 K = K.intersection(W_cone)
205
206 # We've already intersected K with the span of K2, so every
207 # generator of K should belong to W now.
208 K_W_rays = [ W.coordinate_vector(r) for r in K.rays() ]
209
210 L = ToricLattice(W.dimension())
211 return Cone(K_W_rays, lattice=L)
212
213
214
215 def discrete_complementarity_set(K):
216 r"""
217 Compute the discrete complementarity set of this cone.
218
219 The complementarity set of a cone is the set of all orthogonal pairs
220 `(x,s)` such that `x` is in the cone, and `s` is in its dual. The
221 discrete complementarity set is a subset of the complementarity set
222 where `x` and `s` are required to be generators of their respective
223 cones.
224
225 For polyhedral cones, the discrete complementarity set is always
226 finite.
227
228 OUTPUT:
229
230 A list of pairs `(x,s)` such that,
231
232 * Both `x` and `s` are vectors (not rays).
233 * `x` is a generator of this cone.
234 * `s` is a generator of this cone's dual.
235 * `x` and `s` are orthogonal.
236
237 REFERENCES:
238
239 .. [Orlitzky/Gowda] M. Orlitzky and M. S. Gowda. The Lyapunov Rank of an
240 Improper Cone. Work in-progress.
241
242 EXAMPLES:
243
244 The discrete complementarity set of the nonnegative orthant consists
245 of pairs of standard basis vectors::
246
247 sage: K = Cone([(1,0),(0,1)])
248 sage: discrete_complementarity_set(K)
249 [((1, 0), (0, 1)), ((0, 1), (1, 0))]
250
251 If the cone consists of a single ray, the second components of the
252 discrete complementarity set should generate the orthogonal
253 complement of that ray::
254
255 sage: K = Cone([(1,0)])
256 sage: discrete_complementarity_set(K)
257 [((1, 0), (0, 1)), ((1, 0), (0, -1))]
258 sage: K = Cone([(1,0,0)])
259 sage: discrete_complementarity_set(K)
260 [((1, 0, 0), (0, 1, 0)),
261 ((1, 0, 0), (0, -1, 0)),
262 ((1, 0, 0), (0, 0, 1)),
263 ((1, 0, 0), (0, 0, -1))]
264
265 When the cone is the entire space, its dual is the trivial cone, so
266 the discrete complementarity set is empty::
267
268 sage: K = Cone([(1,0),(-1,0),(0,1),(0,-1)])
269 sage: discrete_complementarity_set(K)
270 []
271
272 Likewise when this cone is trivial (its dual is the entire space)::
273
274 sage: L = ToricLattice(0)
275 sage: K = Cone([], ToricLattice(0))
276 sage: discrete_complementarity_set(K)
277 []
278
279 TESTS:
280
281 The complementarity set of the dual can be obtained by switching the
282 components of the complementarity set of the original cone::
283
284 sage: set_random_seed()
285 sage: K1 = random_cone(max_ambient_dim=6)
286 sage: K2 = K1.dual()
287 sage: expected = [(x,s) for (s,x) in discrete_complementarity_set(K2)]
288 sage: actual = discrete_complementarity_set(K1)
289 sage: sorted(actual) == sorted(expected)
290 True
291
292 The pairs in the discrete complementarity set are in fact
293 complementary::
294
295 sage: set_random_seed()
296 sage: K = random_cone(max_ambient_dim=6)
297 sage: dcs = discrete_complementarity_set(K)
298 sage: sum([x.inner_product(s).abs() for (x,s) in dcs])
299 0
300
301 """
302 V = K.lattice().vector_space()
303
304 # Convert rays to vectors so that we can compute inner products.
305 xs = [V(x) for x in K.rays()]
306
307 # We also convert the generators of the dual cone so that we
308 # return pairs of vectors and not (vector, ray) pairs.
309 ss = [V(s) for s in K.dual().rays()]
310
311 return [(x,s) for x in xs for s in ss if x.inner_product(s) == 0]
312
313
314 def LL(K):
315 r"""
316 Compute the space `\mathbf{LL}` of all Lyapunov-like transformations
317 on this cone.
318
319 OUTPUT:
320
321 A list of matrices forming a basis for the space of all
322 Lyapunov-like transformations on the given cone.
323
324 EXAMPLES:
325
326 The trivial cone has no Lyapunov-like transformations::
327
328 sage: L = ToricLattice(0)
329 sage: K = Cone([], lattice=L)
330 sage: LL(K)
331 []
332
333 The Lyapunov-like transformations on the nonnegative orthant are
334 simply diagonal matrices::
335
336 sage: K = Cone([(1,)])
337 sage: LL(K)
338 [[1]]
339
340 sage: K = Cone([(1,0),(0,1)])
341 sage: LL(K)
342 [
343 [1 0] [0 0]
344 [0 0], [0 1]
345 ]
346
347 sage: K = Cone([(1,0,0),(0,1,0),(0,0,1)])
348 sage: LL(K)
349 [
350 [1 0 0] [0 0 0] [0 0 0]
351 [0 0 0] [0 1 0] [0 0 0]
352 [0 0 0], [0 0 0], [0 0 1]
353 ]
354
355 Only the identity matrix is Lyapunov-like on the `L^{3}_{1}` and
356 `L^{3}_{\infty}` cones [Rudolf et al.]_::
357
358 sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
359 sage: LL(L31)
360 [
361 [1 0 0]
362 [0 1 0]
363 [0 0 1]
364 ]
365
366 sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)])
367 sage: LL(L3infty)
368 [
369 [1 0 0]
370 [0 1 0]
371 [0 0 1]
372 ]
373
374 If our cone is the entire space, then every transformation on it is
375 Lyapunov-like::
376
377 sage: K = Cone([(1,0), (-1,0), (0,1), (0,-1)])
378 sage: M = MatrixSpace(QQ,2)
379 sage: M.basis() == LL(K)
380 True
381
382 TESTS:
383
384 The inner product `\left< L\left(x\right), s \right>` is zero for
385 every pair `\left( x,s \right)` in the discrete complementarity set
386 of the cone::
387
388 sage: set_random_seed()
389 sage: K = random_cone(max_ambient_dim=8)
390 sage: C_of_K = discrete_complementarity_set(K)
391 sage: l = [ (L*x).inner_product(s) for (x,s) in C_of_K for L in LL(K) ]
392 sage: sum(map(abs, l))
393 0
394
395 The Lyapunov-like transformations on a cone and its dual are related
396 by transposition, but we're not guaranteed to compute transposed
397 elements of `LL\left( K \right)` as our basis for `LL\left( K^{*}
398 \right)`
399
400 sage: set_random_seed()
401 sage: K = random_cone(max_ambient_dim=8)
402 sage: LL2 = [ L.transpose() for L in LL(K.dual()) ]
403 sage: V = VectorSpace( K.lattice().base_field(), K.lattice_dim()^2)
404 sage: LL1_vecs = [ V(m.list()) for m in LL(K) ]
405 sage: LL2_vecs = [ V(m.list()) for m in LL2 ]
406 sage: V.span(LL1_vecs) == V.span(LL2_vecs)
407 True
408
409 """
410 V = K.lattice().vector_space()
411
412 C_of_K = discrete_complementarity_set(K)
413
414 tensor_products = [ s.tensor_product(x) for (x,s) in C_of_K ]
415
416 # Sage doesn't think matrices are vectors, so we have to convert
417 # our matrices to vectors explicitly before we can figure out how
418 # many are linearly-indepenedent.
419 #
420 # The space W has the same base ring as V, but dimension
421 # dim(V)^2. So it has the same dimension as the space of linear
422 # transformations on V. In other words, it's just the right size
423 # to create an isomorphism between it and our matrices.
424 W = VectorSpace(V.base_ring(), V.dimension()**2)
425
426 # Turn our matrices into long vectors...
427 vectors = [ W(m.list()) for m in tensor_products ]
428
429 # Vector space representation of Lyapunov-like matrices
430 # (i.e. vec(L) where L is Luapunov-like).
431 LL_vector = W.span(vectors).complement()
432
433 # Now construct an ambient MatrixSpace in which to stick our
434 # transformations.
435 M = MatrixSpace(V.base_ring(), V.dimension())
436
437 matrix_basis = [ M(v.list()) for v in LL_vector.basis() ]
438
439 return matrix_basis
440
441
442
443 def lyapunov_rank(K):
444 r"""
445 Compute the Lyapunov rank (or bilinearity rank) of this cone.
446
447 The Lyapunov rank of a cone can be thought of in (mainly) two ways:
448
449 1. The dimension of the Lie algebra of the automorphism group of the
450 cone.
451
452 2. The dimension of the linear space of all Lyapunov-like
453 transformations on the cone.
454
455 INPUT:
456
457 A closed, convex polyhedral cone.
458
459 OUTPUT:
460
461 An integer representing the Lyapunov rank of the cone. If the
462 dimension of the ambient vector space is `n`, then the Lyapunov rank
463 will be between `1` and `n` inclusive; however a rank of `n-1` is
464 not possible (see [Orlitzky/Gowda]_).
465
466 ALGORITHM:
467
468 The codimension formula from the second reference is used. We find
469 all pairs `(x,s)` in the complementarity set of `K` such that `x`
470 and `s` are rays of our cone. It is known that these vectors are
471 sufficient to apply the codimension formula. Once we have all such
472 pairs, we "brute force" the codimension formula by finding all
473 linearly-independent `xs^{T}`.
474
475 REFERENCES:
476
477 .. [Gowda/Tao] M.S. Gowda and J. Tao. On the bilinearity rank of a proper
478 cone and Lyapunov-like transformations, Mathematical Programming, 147
479 (2014) 155-170.
480
481 .. [Orlitzky/Gowda] M. Orlitzky and M. S. Gowda. The Lyapunov Rank of an
482 Improper Cone. Work in-progress.
483
484 .. [Rudolf et al.] G. Rudolf, N. Noyan, D. Papp, and F. Alizadeh, Bilinear
485 optimality constraints for the cone of positive polynomials,
486 Mathematical Programming, Series B, 129 (2011) 5-31.
487
488 EXAMPLES:
489
490 The nonnegative orthant in `\mathbb{R}^{n}` always has rank `n`
491 [Rudolf et al.]_::
492
493 sage: positives = Cone([(1,)])
494 sage: lyapunov_rank(positives)
495 1
496 sage: quadrant = Cone([(1,0), (0,1)])
497 sage: lyapunov_rank(quadrant)
498 2
499 sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)])
500 sage: lyapunov_rank(octant)
501 3
502
503 The full space `\mathbb{R}^{n}` has Lyapunov rank `n^{2}`
504 [Orlitzky/Gowda]_::
505
506 sage: R5 = VectorSpace(QQ, 5)
507 sage: gs = R5.basis() + [ -r for r in R5.basis() ]
508 sage: K = Cone(gs)
509 sage: lyapunov_rank(K)
510 25
511
512 The `L^{3}_{1}` cone is known to have a Lyapunov rank of one
513 [Rudolf et al.]_::
514
515 sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
516 sage: lyapunov_rank(L31)
517 1
518
519 Likewise for the `L^{3}_{\infty}` cone [Rudolf et al.]_::
520
521 sage: L3infty = Cone([(0,1,1), (1,0,1), (0,-1,1), (-1,0,1)])
522 sage: lyapunov_rank(L3infty)
523 1
524
525 A single ray in `n` dimensions should have Lyapunov rank `n^{2} - n
526 + 1` [Orlitzky/Gowda]_::
527
528 sage: K = Cone([(1,0,0,0,0)])
529 sage: lyapunov_rank(K)
530 21
531 sage: K.lattice_dim()**2 - K.lattice_dim() + 1
532 21
533
534 A subspace (of dimension `m`) in `n` dimensions should have a
535 Lyapunov rank of `n^{2} - m\left(n - m)` [Orlitzky/Gowda]_::
536
537 sage: e1 = (1,0,0,0,0)
538 sage: neg_e1 = (-1,0,0,0,0)
539 sage: e2 = (0,1,0,0,0)
540 sage: neg_e2 = (0,-1,0,0,0)
541 sage: z = (0,0,0,0,0)
542 sage: K = Cone([e1, neg_e1, e2, neg_e2, z, z, z])
543 sage: lyapunov_rank(K)
544 19
545 sage: K.lattice_dim()**2 - K.dim()*K.codim()
546 19
547
548 The Lyapunov rank should be additive on a product of proper cones
549 [Rudolf et al.]_::
550
551 sage: L31 = Cone([(1,0,1), (0,-1,1), (-1,0,1), (0,1,1)])
552 sage: octant = Cone([(1,0,0), (0,1,0), (0,0,1)])
553 sage: K = L31.cartesian_product(octant)
554 sage: lyapunov_rank(K) == lyapunov_rank(L31) + lyapunov_rank(octant)
555 True
556
557 Two isomorphic cones should have the same Lyapunov rank [Rudolf et al.]_.
558 The cone ``K`` in the following example is isomorphic to the nonnegative
559 octant in `\mathbb{R}^{3}`::
560
561 sage: K = Cone([(1,2,3), (-1,1,0), (1,0,6)])
562 sage: lyapunov_rank(K)
563 3
564
565 The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K``
566 itself [Rudolf et al.]_::
567
568 sage: K = Cone([(2,2,4), (-1,9,0), (2,0,6)])
569 sage: lyapunov_rank(K) == lyapunov_rank(K.dual())
570 True
571
572 TESTS:
573
574 The Lyapunov rank should be additive on a product of proper cones
575 [Rudolf et al.]_::
576
577 sage: set_random_seed()
578 sage: K1 = random_cone(max_ambient_dim=8,
579 ....: strictly_convex=True,
580 ....: solid=True)
581 sage: K2 = random_cone(max_ambient_dim=8,
582 ....: strictly_convex=True,
583 ....: solid=True)
584 sage: K = K1.cartesian_product(K2)
585 sage: lyapunov_rank(K) == lyapunov_rank(K1) + lyapunov_rank(K2)
586 True
587
588 The Lyapunov rank is invariant under a linear isomorphism
589 [Orlitzky/Gowda]_::
590
591 sage: K1 = random_cone(max_ambient_dim = 8)
592 sage: A = random_matrix(QQ, K1.lattice_dim(), algorithm='unimodular')
593 sage: K2 = Cone( [ A*r for r in K1.rays() ], lattice=K1.lattice())
594 sage: lyapunov_rank(K1) == lyapunov_rank(K2)
595 True
596
597 The dual cone `K^{*}` of ``K`` should have the same Lyapunov rank as ``K``
598 itself [Rudolf et al.]_::
599
600 sage: set_random_seed()
601 sage: K = random_cone(max_ambient_dim=8)
602 sage: lyapunov_rank(K) == lyapunov_rank(K.dual())
603 True
604
605 The Lyapunov rank of a proper polyhedral cone in `n` dimensions can
606 be any number between `1` and `n` inclusive, excluding `n-1`
607 [Gowda/Tao]_. By accident, the `n-1` restriction will hold for the
608 trivial cone in a trivial space as well. However, in zero dimensions,
609 the Lyapunov rank of the trivial cone will be zero::
610
611 sage: set_random_seed()
612 sage: K = random_cone(max_ambient_dim=8,
613 ....: strictly_convex=True,
614 ....: solid=True)
615 sage: b = lyapunov_rank(K)
616 sage: n = K.lattice_dim()
617 sage: (n == 0 or 1 <= b) and b <= n
618 True
619 sage: b == n-1
620 False
621
622 In fact [Orlitzky/Gowda]_, no closed convex polyhedral cone can have
623 Lyapunov rank `n-1` in `n` dimensions::
624
625 sage: set_random_seed()
626 sage: K = random_cone(max_ambient_dim=8)
627 sage: b = lyapunov_rank(K)
628 sage: n = K.lattice_dim()
629 sage: b == n-1
630 False
631
632 The calculation of the Lyapunov rank of an improper cone can be
633 reduced to that of a proper cone [Orlitzky/Gowda]_::
634
635 sage: set_random_seed()
636 sage: K = random_cone(max_ambient_dim=8)
637 sage: actual = lyapunov_rank(K)
638 sage: K_S = _restrict_to_space(K, K.span())
639 sage: K_SP = _restrict_to_space(K_S.dual(), K_S.dual().span()).dual()
640 sage: l = K.lineality()
641 sage: c = K.codim()
642 sage: expected = lyapunov_rank(K_SP) + K.dim()*(l + c) + c**2
643 sage: actual == expected
644 True
645
646 The Lyapunov rank of any cone is just the dimension of ``LL(K)``::
647
648 sage: set_random_seed()
649 sage: K = random_cone(max_ambient_dim=8)
650 sage: lyapunov_rank(K) == len(LL(K))
651 True
652
653 We can make an imperfect cone perfect by adding a slack variable
654 (a Theorem in [Orlitzky/Gowda]_)::
655
656 sage: set_random_seed()
657 sage: K = random_cone(max_ambient_dim=8,
658 ....: strictly_convex=True,
659 ....: solid=True)
660 sage: L = ToricLattice(K.lattice_dim() + 1)
661 sage: K = Cone([ r.list() + [0] for r in K.rays() ], lattice=L)
662 sage: lyapunov_rank(K) >= K.lattice_dim()
663 True
664
665 """
666 beta = 0
667
668 m = K.dim()
669 n = K.lattice_dim()
670 l = K.lineality()
671
672 if m < n:
673 # K is not solid, restrict to its span.
674 K = _restrict_to_space(K, K.span())
675
676 # Non-solid reduction lemma.
677 beta += (n - m)*n
678
679 if l > 0:
680 # K is not pointed, restrict to the span of its dual. Uses a
681 # proposition from our paper, i.e. this is equivalent to K =
682 # _rho(K.dual()).dual().
683 K = _restrict_to_space(K, K.dual().span())
684
685 # Non-pointed reduction lemma.
686 beta += l * m
687
688 beta += len(LL(K))
689 return beta