Orthogonal Splitting: how it relates to the Representation Problem of Quadratic Forms

 As previously mentioned, one of the key questions of the theory of the quadratic space is:

Given an \(R\)-quadratic module \(V\) and \(\alpha \in F = \text{Frac}(R)\), when does \(V\) represents \(\alpha\)? (Written \(\alpha \to V\)). 

1. Orthogonal Splitting

Let \(V\) be a quadratic space over a field \(F\). 

We write \(V = W_1 \perp W_2\) if there are subspaces \(W_i\) such that \(V = W_1 \oplus W_2\) and every element of \(W_1\) is orthogonal to every element of \(W_2\). 

Given a subspace \(W\) of \(V\), we define its orthogonal complement \(W^{\perp} \) to be the set of vectors in \(V\) orthogonal to every vector in \(W\). 

Question. Is it true that if  \(V = W \perp U\) then \( U = W^{\perp}\)?

Answer. No because \(W^{\perp} \cap W\) might be nonzero so the sum \( W + W^{\perp}\) might not be direct. 

 We call the set \( \text{rad}(W):= W \cap W^{\perp}\) the radical of \(W\). Thus the radical of \(W\) consists of \(w \in W\) that are orthogonal to every element of \(W\).  If \( \text{rad}(W) = \{0\} \) then \(W\) is said to be regular. Above we see that if \(W\) is NOT regular then \(V \neq W \perp W^{\perp}\). It turns out that this is the only obstruction. 

Before showing that this is the case, let us find another characterization of regularity. 

1.1. Regularity in terms of Discriminant.

 Recall every quadratic module over a field has an orthogonal basis. So suppose \(V \cong \langle \alpha_i, \ldots, \alpha_n\rangle\) in some  orthogonal basis  \(\{e_1, \ldots, e_n\} \). Recall that this means \( B(e_i, e_i) = \alpha_i\) and the Gram matrix of \(V\) with respect to \( \{e_i\}\) is therefore 

\[A = \begin{pmatrix}\alpha_1 & 0 & \cdots & 0 \\ 0 & \alpha_2 & \cdots & 0 \\ 0 & 0 & \cdots &\alpha_n \end{pmatrix}. \]

In particular \(d V \equiv \det A = \prod_i \alpha_i \).

Now for every vector \(x = \sum c_i e_i \in V\) we have \(B(x, e_i) = c_i B(e_i, e_i).\)  So \[x \in \text{rad}(V) \iff B(x, e_i) = 0 \,\,\forall \, i \iff c_i B(e_i, e_i) = c_i \alpha_i= 0 \,\,\,\, \forall \, i.\]

Thus we have \[\text{rad}(V) \neq \{0\} \iff \exists c_i \neq 0 \text{ s.t } c_i \alpha_i = 0 \iff \text{ one of the } \alpha_i \text{ is } 0 \iff \det A = 0. \]

Thus we have established the following.

Prop. A quadratic \(F\)-space \(V\) is regular \(\iff\) the determinant of its Gram matrix is nonzero.

Remark: We will often abuse notation and write the second condition as \(dV \neq 0\) though strictly speaking \(dV \) is the equivalence class of \(\det A\) in \(F^*/(F^*)^2\). 

 

1.2. Regular Subspaces Split. 

For regular subspaces, Orthogonal Complements are truly Complements!

Theorem. Let \(W\) be a regular subspace of \(V\). Then we have  

•  \(V = W \perp W^{\perp}\). 
• Furthermore, if \(V = W \perp U\) then \(U = W^{\perp}.\) 

Proof of Theorem. 

• From the above discussion, we already see that \(W \cap W^{\perp} = \{0\} \).  It remains to show that \(W + W^{\perp} = V\). Let \(x \in V\), we want to show that there exists \(w \in W\) such that \((x - w) \in W^{\perp}\), i.e. \(B(x - w, - )|_W \equiv 0\). Or equivalently \(B(x,-) = B(w, -)\) on \(W\). 

Since \(B(x, -)|_W \in W^*\), it suffices for us to show that every linear functional in \(W^*\) is of the form \(B(w, -)\) for some \(w \in W\). 

Prop. Define a map \(f: W \to W^*\) by \(w \mapsto B(w, -)\). Then \(f\) is an isomorphism \( \iff W \) is regular. 

proof of the proposition. Fix a basis \(\{e_1, \ldots, e_n\}\) of \(W\) and let \(A = (a_{ij}) = \left( B(e_i, e_j)\right)\) be the corresponding Gram matrix of \(W\). Then we claim that the matrix of \(f\) with respect to the bases \( \{e_i\}, \{e_i^*\}\) is \(A\) 

. 

Indeed, let's compute \(f(e_i) \in W^*\). We have \(f(e_i)(e_j) = B(e_i, e_j) = a_{ij}\), so \(f(e_i) = \sum_j a_{ij} e_j^*\). 

Thus \(f\) is an isomorphism iff \(A\) is invertible. The later holds iff \( \det A \neq 0\), and by the previous section on regularity, that is equivalent to \(W\) being regular. \(\square\)

•  Now, finally suppose \(V = W \perp U \) then \(U \subset W^{\perp}\) and \( \dim U = \dim V - \dim W = \dim W^{\perp}\). Therefore \(U = W^{\perp}\). \(\blacksquare\)

2. Relation to Representations.

Corollary. Let \( \alpha \in F^*\). Then \[ \alpha \to V \iff \text{ there is a splitting } V = \langle \alpha \rangle \perp U .\]

Proof. Suppose \( q(w) = \alpha\) for some \(w \in V\). Let \(W = \text{span}_F(v) \subset V\). Then \(W\) is clearly regular: a vector \((c w) \in W \) lies in its radical \(\iff\) \(c \alpha = B(cw, w) = 0\), i.e. \(c = 0\).  Thus by the Theorem in Section 1.2, we must have an orthogonal splitting \(V = W \perp W^{\perp} = \langle \alpha\rangle \perp W^{\perp}.\) 

Conversely suppose \(V = \langle \alpha \rangle \perp U \). This means in particular that \(V\) contains a subspace \( W \cong \langle \alpha \rangle\) in some basis \(\{w\}\) of \(W\). Thus \(q(w) = \alpha\). \(\blacksquare\) 

Three easy ways to distinguish quadratic forms: Orthogonal Basis, Discriminant and number of Representations.

 When talking about representations of quadratic forms, we need to specify its domain. 

Example. Question: can the form \(q(x) = x^2 \) represents \(-1\)? 

Answer. It depends, over \(\mathbb{C}\) yes, but not over \( \mathbb{R}\). 

This motivates us to change our viewpoint from studying quadratic forms as just some polynomial expressions to studying \(R\)-modules equipped with a quadratic form.  We call such modules quadratic \(R\)-modules. Two such modules \(M, N\) are called isometric if there exist a bijection \( \varphi: M \to N\) that preserves the quadratic form, i.e. \( q_M(x, y) = q_n(\varphi(x), \varphi(y) \) for all \(x, y \in M \). 

Below we discuss three elementary ways to check if two quadratic modules are isometric. 

1. Orthogonal Basis 

A useful tool in studying quadratic \(R\)-modules is its orthogonal basis. Every quadratic form is induced from a unique bilinear form B(x,y). Using B(x, y), we can define the notion of orthogonality in an obvious way. The question is: Does every quadratic \(R\)-module have an orthogonal basis?

Answer: No. Remember in Linear Algebra, we produce orthogonal basis via Gram-Schmidt, which involves division by norms of vectors. So to be able to reproduce Gram-Schmidt, we need to be able to divide by elements of \(R\)- i.e. we want \(R\) to be a field. Indeed we have the following theorem.

Theorem. If V is a quadratic \(R\)-module over a field \(R = F\) (in which case we call it a quadratic \(F\)-space), then V has an orthogonal basis \(\{e_1, \ldots, e_n\}\). If \(B(e_i, e_i) = \alpha_i \), then we write \(V \cong \langle \alpha_1, \ldots, \alpha_n \rangle \) in \( \{e_1, \ldots, e_n\}\). 

Proof. The proof essentially use  Gram-Schmidt with one caveat: some non-zero element \(v\)  of \(V\) might have zero "norm", \(q(v) = 0\). (We call such a \(v\) isotropic.)

Now scaling an orthonormal basis gives an orthonormal basis. As \(B(c e_i, c e_i) = c^2 B (e_i, e_i)\) we see that if  \(V \cong \langle \alpha_1, \ldots, \alpha_n \rangle \) in \( \{e_1, \ldots, e_n\}\) then  \(V \cong \langle \alpha_1/c_1^2, \ldots, \alpha_n/c_n^2 \rangle \) in \( \{e_1/c_1, \ldots, e_n/c_n\}\). In other words, the tuples \( \alpha_i \) are only determined up to a square in \(F^*\). 

Example. Are the spaces \( \langle 1, 2\rangle, \langle 2, 2\rangle\) and \( \langle 2, 3\rangle\) isometric over \( \mathbb{R}\)?

Answer: Yes, because by the above discussion, they are all isometric to \( \langle 1, 1\rangle\). For example if \(V \cong \langle 2, 3\rangle \) in \(\{e_1, e_2\}\) then \(V \cong \langle 1, 1\rangle \) in \( \{e_1/\sqrt{2}, e_2/\sqrt{3}\}\). 

More generally, observe that over \(\mathbb{R}\) if \( \alpha_i / \beta_i \geq 0 \) and is thus a square in \( \mathbb{R}*\) then \( \langle \alpha_1, \ldots, \alpha_n \rangle \cong \langle \beta_1, \ldots, \beta_n\rangle \). In particular, every quadratic space over \(\mathbb{R}\) is of the form \( \langle \alpha_1, \ldots, \alpha_m, \beta_1, \ldots, \beta_n\rangle \) for \( \alpha_i \geq 0, \beta_i < 0\). Thus every quadratic space over \( \mathbb{R}\) is isometric to one of form \(\langle 1, \ldots, 1, - 1, \ldots -1 \rangle\). 

Even more generally, we have the following.

Lemma. Let \(F\) be a field. Suppose a full set of coset representatives of \(F^*/(F^*)^2\) is \(\{ \epsilon_1= 1, \ldots, \epsilon_n\}\). Then  every quadratic space over \(F\) is isometric to one of form \[ \langle \epsilon_1 =1, \epsilon_1, \ldots, \epsilon_1, \epsilon_2, \ldots, \epsilon_2, \ldots, \epsilon_n, \ldots, \epsilon_n\rangle. \]

Example. Since every element of \( \mathbb{C}^*\) is a square, every quadratic space over \( \mathbb{C}\) is isometric to one of form \( \langle 1, 1, \ldots, 1 \rangle \).

Example. Let \(F = \mathbb{F}_q\) be a finite field. Since the kernel of the square map on \( F^*\) is \( \{\pm 1\} \), we must have its image, the group of squares \( (F^*)^2\), must be of side \(|F^*|/2\), i.e. it is a subgroup of index \(2\). Thus every quadratic space over a finite field \(F\) is isometric to one of form \(\langle 1, 1, \ldots, 1, \epsilon, \ldots, \epsilon\rangle\) for some nonsquare \(\epsilon \in F^*\).   

 

2. Discriminant

To each quadratic \(R\)-module \( (M, q, B) \) with a fixed basis \( \{v_1, \ldots, v_n\} \) we can associate a Gram matrix \(A = \left(B(v_i, v_j) \right) \). We write \(M \cong A\) in \( \{ v_1, \ldots, v_n \} \). If we change basis then the new matrix \(B\) will be congruent to \(A\), i.e. \(B = T^t A T\) for some invertible matrix \(T\). Thus \(\det B = (\det T)^2 \det A \), so the determinant of the Gram matrix of \(M\) is uniquely determined up to a square in \(R^*\). We define the discriminant of \(M\), denoted \(dM\) to be the image of \(\det A\) in \(R^*/(R^*)^2\).

How does the discriminant help with classification? Well, for one thing, if \(dM \neq dN\), then \(M \not \cong N\). 

Example. The \(\mathbb{Q}\)-modules \(  M \cong \langle 1, 1 \rangle\) and \(N \cong \langle 2, 3 \rangle \) are not isometric over \( \mathbb{Q}\) since  \( dM = 1 \not \equiv 6 = dN \mod (\mathbb{Q}^*)^*\). 

Note that above we showed \(M \cong N\) over \(\mathbb{R}\) and indeed their discriminant differ by a square in \( \mathbb{R}^*\): \(dN = (\sqrt{6})^2 dM \equiv dM\) in \( (\mathbb{R}/\mathbb{R}^*)^2.\)

3. Number of Representations

Example. Are the following \( \mathbb{Z} \)-modules isometric?

\[ M \cong \begin{pmatrix}2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 5 \end{pmatrix}; \]

and

\[ N \cong \begin{pmatrix}2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{pmatrix}; \]

Answer. Since \(dM = dN = 13 \) we cannot use discriminant to distinguish them. However we observe that \(M\) represents \(2\) exactly 6 times while \(N\) represents \(2\) only twice. So they cannot be isometric.

Here we have used the following idea.

If the quadratic \(R\)-modules \(M, N\) are isometric, then for every \(r \in R\), we have a bijection \[ \{ m \in M \mid q_M(m) = r\} \leftrightarrow \{ n \in N \mid q_N(n) = r\}. \]

Quadratic Forms

An important question in Number Theory is:  which primes can be written as \(x^2 + n y^2\)? This question can be thought of in many different ways: in terms of factorization of primes in algebraic extensions, in terms of norms of number fields but the original viewpoint of Gauss in Disquisionaire Mathematics is in terms of Quadratic Forms.

In the language of Quadratic Forms, the question becomes, which primes can be represented by the quadratic form \(q(x,y) = x^2 + ny^2\). 

The theory of Quadratic Forms revolves around two fundamental problems.

• Representation: Which "numbers" can be represented by a given quadratic form q? 
• Classification: When are two forms "equivalent"? (i.e. can be obtained from each other via an invertible change of variable). 

Key idea of Pete Clark's paper on Elementary Equivalence versus Isomorphisms

 From Pop, we know that if \(K\) and \(L\) are finitely generated fields (i.e. over their prime subfields) and \( K \equiv L\) then there exist embeddings \(K \hookrightarrow L\) and \(L \hookrightarrow K\). In other words, \(K\) and \(L\) are isogeneous.  

Thus we want to see when being isogenous implies being isomorphic. Clark shows that such would be the case if we have one of the following three situations:

1. For some positive integer \(n\),  \(K\) and \(L\)  are cyclic elements of \(SB_n\), the set of function fields of Severi-Brauer varieties of dimension \(n\) over a given field \K\).

2. For some \(n \leq 2\), both \(K\) and \(L\) lie in \(Q_n\), the class of function fields of quadric hypersurfaces of dimension \(n\) over \(k\) and \( char \,k \neq 2\).

3. For some \(n > 1\), \(K \in SB_n\) and \(L \in Q_n\). 

Question: does elementarily equivalence always \( \implies\) isogeny? Observe that Pop's theorem was only for finitely generated function fields. 

Answer: If \(K \equiv L\) then their Brauer kernels are going to be Galois conjugates. Thus for any choice of \(k\)-structure on \(K\), ther is a unique \(k\) structure on \(L\) such that the Brauer kernels \(\kappa(K/k)\) and \(\kappa(L/k)\) are equal.  What Clark proof is that this is basically equivalent to \(K\) and \(L\) being isogenous, as well as to \(K\) being isomorphic to \(L\). 

The key player here is the Brauer kernel and the main idea use is the relevance of Galois cohomology and Brauer groups to the classification of Quadric surfaces. 

Remember this plan

 I’m worried about my parents’ health so I will devise a good program and write a book for them. I want to at the same time, create some programs and books for Vietnamese kids. I have many education plans in English, but then I realize there are already so many resources for English speakers- I should do something for my country instead. Let’s remember to do this. It will be useful for my future children too since I will lean how to teach them in Vietnamese instead. It’s a good chance for me to review Vietnamese and not forget my language. 

Inducing autonomous/undirected behavior

I often worry about my future children falling victim to the curse of the Western modern life, wasting their life away sitting around watching TV, surfing social media and getting mindlessly engrossed in other soul sucking activities. I worry that they will get addicted to eating unhealthy process food, to sitting, to staying in their comfort zone, to passive entertainment. How do I get them to love learning, thinking, challenges, physical activities  and just growing in general?

When I was younger, I often fantasized about having some kind of tiger mom regime for my kids. Well, thank God I didn’t have kids at 16. Those regimes will create passive compliance and kill any personal inner drive and initiative. I was so foolish to want a system that would need my supervision 24/7. Wouldn’t it be nicer to have a autonomous system- I just set up some infrastructures or rule and then off they go on their own (mostly, not completely, like a mixed economy)? 

As I grew older, I have become more and more libertarian, not on political matter, but rather, on issues regarding my own mental physical health. That is, my body, with its impulses, reflexes and conditioning, is like a country. I constantly ponder how to manage and govern it. When I was younger, I was too much of a helicopter parent and a micromanager of my body. Now I learn when to step back (emphasis on when- not always. I’m anti-extremism.) and delegate tasks to the unconscious and the Lower brain. It definitely helps so much, not just with posture, dancing, exercising, but also with creativity, thinking, resilience and empathy (with others and myself). 

However, one can’t simply just leave a system to itself and expect everything to work out fine. Try just leaving any students, kids, economy, teams completely unsupervised. The art of lassaifairing is not as simple as it sounds. First one has to trigger the system- remember the law of inertia? One has to create a momentum first, otherwise nothing will run. There are many ways of triggering such motion, e.g. priming and conditioning are some basic examples. But today I want to talk about triggering via pleasure.

The reason I like exercises, stretching, Alexander techniques or dancing is not because I know that they are good for me. I don’t do them purely because of a sense of responsibility or discipline. Rather, I derive great joy and pleasure from them. For exercise, yoga, qigonhg and general stretching, a large part of the pleasure comes from eradication of tension, pain and sluggishness. So actually having those negative sensations was a good thing- it intensifies the pleasure I have when I exercise. Once I have experienced the consequences of not exercising and notices the change when I exercise, it naturally leads my body to crave exercising. It associates exercising with that positive sensation, thus triggering dopamine release at the anticipation of exercising even. This is analogous to the case of overeating- the method that both I and Milton Erickson use to reduce craving and addiction to bad food. It creates a positive feeling associated with not having the food. I have been adopting this method “too much pain to the point of letting go” in life since that crazy arguments we had at Conic in 2007 with my roommates. The release I felt then was magical. So whenever things go bad these days, I actually know that once enough pain accumulated, it will force me to let go, thus creating a wonderful release. So a way of freeing myself from attachments to certain ideas and people is to instead of forbidding it, letting myself be exposed so much that the pain become overwhelming. 

In terms of parenting, I should create situations in which the kids can actually feel the pleasure from being resilient, from physical activities, from exploring or from meeting challenges. I should set up activities in which kids can derive pleasure and see the pay off from grit. That internal reinforcement will condition them to build a habit that won’t need will power or my control to maintain. In a sense, I am trying to establish/calibrate a good utility function for them- one that assigns high value to the right things. 

“Grow taller”

 I finally understood “lengthen your body”, “get longer”, “grow taller”, “think up”. I guess not 100%, but I have made some breakthrough! So my tango teachers have always told me to lengthen my spine, but upon hearing that, I always just arched my back or my neck backwards instead. Now slowly from Alexander techniques, Qi gong and stretching, I understood the feeling of pulling my limbs out. By understanding I mean, I can sense it the moment it happens and once I know how that sensation should feel like, I was able to encourage it and even make it happen. I guess I bring it under my conscious command and thus gain some control over it. I only realize now that in the past, when doing forward fold and dangling, I have actually been tensing and restricting my movement. Once I have learned how to consciously create this sensation of letting go, the forward fold actually pull my upper body out of my lower body, thus lengthening my height. It felt magical. The same thing happens with other stretches. I guess I have been stretching incorrectly, resisting the stretching instead of letting it pull my body out and create space?

In bringing things into conscious awareness, I have created a button, a knob, a controller, a toolbar, to be able to adjust the specific sensation. 

Now though, I need a way to maintain the relaxation and space. When writing and reading, I often unconsciously fall back to my bad habits of gripping, tensing, contracting and thus create a lot of pain.