Well, I begin to write something while I’m reading. I hope it can be more productive than before. Anyway, let’s begin. I’m now reading Qianchu Yuan’s post “The Jacobi-Trudi identities” right now. I’m lazy enough to make absolutely no change to the title. The title is not quite informative though. I don’t understand what Schur function is, but this post is going to introduce one definition of it. Let be a patition. What’s patition by the way? , , weight , sum over all tableaux of shape . Verify , . What’s and by the way? I can’t follow this post anymore, perhaps I should check the previous post first. Ok, now I’m back to Qiaochu Yuan’s post Young diagrams, q-analogues, and one of my favorite proofs, which at first sight, seem accessible by me. Young diagrams is a visual representation of partition, that reminds me of the definition. I think I have seen it before. A partition of is a sequence such that . Young diagram consists of rows and row has boxes. Let denote the poset of Young diagrams which fit into an box, ordered by inclusion. Verify . Mmm, let me think, well, that’s an obvious combinotorial exercise. I have done it hundreds of times during my middle school years. So let’s move on. Now what’s -analogue. Let’s be a bit patient.We now consider the number of boxes, which we denote . Wait, what do we mean by the number of boxes, let me check. Let us first define and we have the result . Let me play around it for a while. Mmm, that means and what does that mean? I can’t see it still. Let me look at some other places. Now after checking Wikipaedia, I think the number boxes in is $m\times n$(The reason that I’m previous confused is Qiaochu said $m\times n$ box in several lines before which really obscures the meaning of box). Then it seems wrong that . Anyway, let’s move on and check back later. Oh, I suddenly see. Every power of 1 is 1, that’s why . By flipping the diagonal, we get . We have the property and . The first one is obvious. The second one is straight forward too. COnsider the first row of a Young diagram in . If it’s not of length , it’s an element of . Otherwise we delete it, then it’s in . The accounts for the deletion(since the first row has boxes in this case which further consolidates my belief in what “box” is). Define , since $R_1(n-1,1)=n$, this can be viewed as the -analogue of the number . Define and . We wish to prove . But for now let’s forget all about the computation and focus on the conception. We have the proposition: If is a power of a prime, then is the number of -dimensional subspaces of . Now consider . . Wait, I think there’s something wrong here. Shouldn’t it be ? Anyway, let’s temporarily jump this. acts transitively on of -dimensional subspaces of , this is Grassmannian by the way. I first encountered this in my Differential Topology class. It’s already a year ago. So from the orbit-stabilizer theorem(a theorem which quite fits my intuition, I like that), we get where denotes the stabilizer subgroup of any particular -dimensional subspace. So we compute the stabilizer subgroup of the space spanned by the first $k$ standard basis vector: any stabilizer is block upper-triangular with block consists an element of and columns chosen arbitrarily. So (lovely combinatorics) there’s . So . After all, will be canceled out and it doesn’t really matter. But what it has to do with Young diagrams? They are related by the so-called “cellular decomposition” of the Grassmannian. The rest omitted, but I have understood the correspondence at this point. Now let’s look at if there are other posts about Young diagram.

And we do find another post Standard Young tableaux and Robinson-Schensted-Knuth. A Young tableau is a chain in Young’s lattice which can be represented in a augumenting way. Given a Young diagram , let denote the number of Young tableaux of shape (I already see why it’s not unique). Another notation, if is the partition of , we write or . Theorem . Well, the proof ideas is to show a bijection between permutations and pairs of standard Young tableaux of the same shape. More details are not explicit in the post.

Now finally we are back to The Jacobi-Trudi identities again. This time I did see semistandard Young tableau differs from standard ones in the way that integers are only weakly increasing along rows. Now I guess means a Young diagram. Equivalent definitions: , . So there exists one problem, I still don’t get what and is. Let me check it out. After checking wiki, is complete homogeneous symmetric polynomials and is elementary symmetric polynomials. Ah, now I understand it. Let me play around why , . Umm, that’s basically because the integers weakly increase along the row and strictly increase along the column, yeah, I see the reason. Let’s move on the proof of equivalence of definitions. While reading the proof, I suddenly found the $\latex det$ part is related to Qiaochu’s previous post The many faces of Schur functions.

So now we are on another post again. It’s quite frustrating to going backward and forth, I really hate that as you do. Let’s see what we’ve got. Ah, there’s a confliction again. perhaps means exterior algebra. Anyway, I’m tired, perhaps I’ll pick up this post at some future time. Have a good day.

It has been my long-standing curiosity about the reference on the inequality related to Riemann Hypothesis. Today accidentally I found it. It’s in “Sharper Bounds for the Chebyshev Functions and . II”, page 339, Corollary 1

The reduction is from Ising spin glasses. This post is an excerpt from Schuch’s paper with my comments.

First we should check Hartree-Fock is in NP, to do this we notice the ground state is . And since the determinant can be computed in polynomial time, we can efficiently evaluate whether a ground state is below a threshold. Another way of writing the ground state is where and $latex |\Omega\rangle$ is the vacuum. Note that $b_i$ is a fermion creater which is antisymmetric and satisfy is 0. So it’s equivalent to the determinant expression.

Basically what we do is add to penalize double occupancy and convert to . Only one of the four terms actually appear since we add the penalty for double occupancy previously. represents the sign of the product . The ground state mapping is to and to