My general feeling at the end of Polymath11 has been that we had just scratched the surface of the problem, getting some feel for how difficult it is. The naive hope that I had at the end is that it may be possible to derive a constant lower bound (but smaller than ) by proving a suitable result on approximating every finite lattice by another type of lattice for which FUNC is known to hold, e.g. a geometric lattice. This goes in a vaguely similar direction as what you’re suggesting. But I don’t know how to go about this technically, and it will definitely require an enormous effort.

In any case, perhaps this discussion would be more appropriate over at the last Polymath11 thread?

]]>1. Frankl’s Conjecture asserts that in any family of sets (1,2,3…) that is union closed, there is a least one element (a,b,c…) found in at least half of all the subsets of the set containing all elements in the family.

2. Arrange the sets in ascending order of cardinality, from smallest set to the set containing all elements. Sets of equal cardinality can be put in any order within the overall arrangement.

3. The smallest set is “irreducible.” The union of any two sets is either a subset of the larger one, or has cardinality greater than the larger one, or is, in the case of two sets of equal cardinality, a set of greater cardinality.

4. Number the sets from 1 to N in ascending order, N being the set of greatest cardinality.

5. Create the power set of all sets, 1 to N – 1.

6. Because set union is associative and commutative, and because union-closed families include every union, every set represented in the power set is a member of the union closed family, one of the numbered sets used as generators.

7. Some of these sets are entirely contained in their set of greatest cardinality. Some include all the sets enumerated, but belong to a set of greater cardinality that is not part of the bracketed sets.

8. Because the number of subsets of the power set is strictly larger that the set containing the original elements (and grows exponentially), each element will be represented by multiple subsets of the power set.

9. Therefore multiple subsets will be equivalent, representing the same unique generating set.

10. Equivalence of these subsets of the original set implies intersections, i.e., element(s) (a,b,c…) appear in more than one generating set.

11. Using the ‘irreducibility” of the least set (1), you should be able to establish a chain of subsets in which one is included in more than half the original sets (1,2,3…). Obviously, the power set grows must faster than the number of original sets, so a kind of induction might be possible.

12. For example, establish that 1 is a proper subset of 3, establish that 3 is a proper subset of 6, etc. ]]>

Let be a group equipped with a homogeneous norm ( is necessarily abelian and torsion-free). Then this norm extends to a norm on the -vectorspace . In detail, every element of has a representative of the form with and we set .

Now let $\overline{A_\mathbb{Q}}$ be the metric completion of . This is still a normed group into which embeds, and is in fact an -vectorspace: if and are Cauchy sequences and then so is and it is easy to check that its equivalence class depends only on the equivalence classes on the original sequences. Since also it also follows that the norm is -homogenous. In summary, is naturally a complete normed -vector space, that is a Banach space.

Finally, there is a natural map of -vectorspaces . This map is injective since our extension of the norm was still a norm. Since the RHS is compatibly an -vectorspace, this induces a further map of -vectorspaces

.

However, the latter map need not be injective! In other words, when we pull back the norm from to , the result need only be a seminorm.

For example, if we start with the norm on then the same formula defines a norm on but only a seminorm on . To get a norm we need to divide by the subspace which is compatible with everything we’ve said since this subspace is disjoint from the image of here so and still inject in the quotient (isometrically, as they must).

]]>arbitrary elements). ]]>

,

assuming that satisfies homogeneity and the triangle inequality on the nose. Thus .

Doing a more refined analysis along the lines of Proposition 1 in the second blog post could be interesting, but I haven’t done this yet and find it hard to say whether it would have the potential to lead to an alternative proof or not.

]]>this feels rather late in the day (or sometime the next day) but

still: why can’t we argue thus….

Given with homogeneous (pseudo-) length function ,

let be your favourite upper bound that holds for all

commutators, thus for all . By

Culler’s identity

(sometimes seen in introductions to stable commutator length)

we have , thus we now know holds for any

commutator. So now use the argument recursively to get

for all and hence .

There were two limitations of the way the computer proof was done:

* While the use of conjugacy invariance and triangle inequality was optimal and algorithmic, of which elements to take powers was manually specified by me. This should have been made smart, and would have soon enough except the extreme smartness of the people in this polymath group made this redundant (problem was solved within 24 hours of the first posted computer proof).

* More importantly, I used [domain specific foundations](https://github.com/siddhartha-gadgil/Superficial/blob/master/src/main/scala/freegroups/LinNormBound.scala), which could encode only one kind of proof, that a specific word has length bounded by a specific number. This rules out in particular both firmulas for bounds that are quatified (and so must invlove variables) and recursion/induction. To show that such results can be at least _encoded_ I formalized the [internal repetition trick](http://siddhartha-gadgil.github.io/ProvingGround/tuts/LengthFunctions.html).

More generally, where a computer helped was in following instructions of the form “try these method in lots of cases in lots of ways and give me the best proof for thess cases (or where we got a strong result)”. It is obvious that the “lots of cases” and “lots of ways” are much bigger numbers for computers than by hand. The question is how general one can be with “these methods”. I do think even in practice a lot of methods can be encoded, and this is underutilized as people underestimate this. In principle, in the era of Homotopy type theory and Deep learning, presumably every method can be encoded.

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