Pure intonation vs. 12 equal temperament musical intervals

Posted on March 26, 2025 by The author

Given a musical sound $X$ , its pitch is characterized by the frequency $f(X)$ of its sound wave. For example, if we regard the pitch as a musical note, then the pitch of frequency $2\times\mathrm{freq}(X)$ corresponds to what is called an octave higher above $X$ ; so we have had this feeling of consonance in between two frequencies differing by a multiple of $2$ . Now a thesis here is that our minds’ ears are musically interested in the ratios of frequencies, and this is the basis of the usual octave division in western music: if you are going to divide pitches between the note $X$ and its higher octave into $12$ notes, then it might be better to put this notes in a geometric progression, i.e. the $k$ -th note has frequency $\mathrm{freq}(X) \times 2^{\frac{k-1}{12}}$ . In this manner, the ratio between any two consecutive notes is equal to a universal amount, $2^{1/12}$ . Pianos, classical guitars, saxophones, and other instruments all obey this tuning, known as 12 equal temperament (12-ET) tuning.

Yet, in the meantime you may have heard of the consonance quality of the perfect fifth interval, e.g. when you play the notes $G$ and $C$ of an octave, that the note $G$ in an octave has the frequency $\frac{3}{2}\times \mathrm{freq}(C)$ for the note $C$ in the same octave, and it’s said that it is because of this simplest ratio of $\frac{3}{2}$ in between them that the two notes sound so resonant to our ears (I checked with Google Gemini now, it said it too). But is the ratio actually equal to $\frac{3}{2}$ ? Doing the math from the above paragraph, we have that the perfect fifth interval $G$ is $8$ divisions (called semitones officially) higher than $C$ , we have $k=8$ in the above formula and

$\mathrm{freq}(G)=\mathrm{freq}(C)\times 2^{7/12} \simeq \mathrm{freq}(C) \times 1.498$

so the actual ratio is not $1.5$ ! What’s happening here?

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Scalar potential and differential forms

Posted on February 22, 2025 by The author

Given a Riemannian manifold $(M^3, g)$ , for a vector field $X$ over $M$ we can define the generalized curl $\nabla \times X = \star (dX)$ where $dX$ is the exterior derivative of (the one-form corresponding to) $X$ and $\star$ is the Hodge star operator with regards to the Riemannian volume form. So for example the vector calculus identity $\nabla \times \nabla f = 0$ in this language becomes $\star d(df)=0$ which follows from $d^2 = 0$ . Similarly we can define the divergence $\nabla \cdot X = \star d( \star X)$ and again the identity “divergence of curl is zero” translates to the equality $\star \ d \star (\star dX) =0$ , which again holds because here we have $\star \star = \mathrm{Id}$ .

In the physical setting, we may have a vector field $E$ over $M$ and a Killing field $\eta$ (i.e. a field with $\mathcal{L}_{\eta} g = 0$ , which will be a Lorentz transformation), and we ask whether there exists a scalar potential for it, meaning that a function $f \in C^\infty(M)$ with $\iota_{\eta}(\star E) = df$ . And what physicists do to test for existence of such an $f$ is to take the curl of the vector field $\iota_{\eta}(\star E)$ and check if it’s zero. But we now see how such a check is justifiable, because $\iota_{\eta}(\star E)$ being curl-free corresponds to it being closed, and in $\mathbb{R}^4$ spacetime closed forms are exact, i.e. potential fields will exist.

All the coefficients of the characteristic polynomial are some traces

Posted on January 30, 2025 by The author

If $V$ is an $n$ -dimensional vector space and $A$ is an matrix over $V$ , you probably remember that in its characteristic polynomial $p_{A}(t) =\det(A-tI)$ , the coefficient of $t^{n-1}$ was minus the trace $\mathrm{Tr}(A)$ and that the constant coefficient was $\det(A)$ . Probably an absurd question might be that, does there exist a formalism and computing principle that produces the coefficient of each $t^k$ , and so $\mathrm{Tr}(A)$ and $\det(A)$ are just the two extremities of the formalism?

And despite the absurdity in the question, I like the surprise in the answer: Yes. There is a generalized notion of the trace map defined for tensors. Note that we have $\bigwedge^1A =A: V \to V$ and $\mathrm{Tr}(\bigwedge^1A)=\sum_{i=1}^n \langle A(e_i), e_i \rangle$ . Then if you let $\bigwedge^kA: \bigwedge^k V \to \bigwedge^k V$ given by $v_{1} \wedge \dots \wedge v_k \mapsto A(v_1) \wedge \dots \wedge A(v_k)$ for all $v_i \in V$ (the notation $A(e_I)$ is usually used to denote $A(e_{i_1}) \wedge \dots \wedge A(e_{i_k})$ where $I = \{ i_{1}, \dots , i_k \}$ is a multi-index of order $k$ ), the trace of $\bigwedge^kA$ is defined as

$\displaystyle \sum_{\text{All increasing } I \text{ with } \lvert I \rvert=k} \langle A(e_I), e_I \rangle$

where $\langle A(e_I), e_I \rangle$ means the evaluation of the $k$ -form $A(e_I)$ on the $k$ -vector $e_I$ .

The similarity of the last formula with that of primary trace is noticeable. It’s like that in the original trace definition, the indices are expanding in one dimension (the index was $i$ enumerating only numbers), so the space of possible pairings is $(1+1)$ -dimensional and we were only summing over in the $1$ dimensional diagonal subspace $\{ (i,i) \}_{1 \leq i \leq n}$ of the index space. Here however the indices are $k$ -dimensional so $2k$ -dimensional for the space of possible pairings; yet here as well we only sum over pairs on the $k$ -dimensional diagonal subspace $\{ (I , I) \}_{\lvert I \rvert=k}$ . For $k=n$ we have $\mathrm{Tr}(\bigwedge^nA)=\langle A (e_1) \wedge \dots \wedge A(e_n), (e_1, \dots, e_n) \rangle = \det(A)$ , and if you stare at the definitions for long enough you can easily verify that

$\displaystyle p_{A}(t)= \sum_{k=0}^n t^{n-k} \cdot (-1)^k \mathrm{Tr}\left(\bigwedge ^k A\right)$

It seems that exterior algebra is a more convenient framework for working with the combinatorial attributes of a matrix, since the alternativity properties needed to define those attributes are already built into wedge products. Let me just mention one other example. The Pfaffian $\mathrm{Pf}(A)$ when $A$ is a $2n \times 2n$ skew-symmetric matrix (not relevant or interesting in other cases) is a useful quantity in geometry and mathematical physics and it’s defined by

$\displaystyle \mathrm{Pf}(A) = \frac{1}{2^n n!} \sum_{\sigma \in S_{2n}} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{\sigma(2i-1),\sigma(2i)}$

but if we let $\omega=\frac{1}{2}\sum a_{ij} e_i \wedge e_j$ , then $\mathrm{Pf}(A)$ can also be retrieved from the equality

$\displaystyle \underbrace{ \omega \wedge \omega \wedge \dots \wedge \omega }_{ n \text{ times} }= \mathrm{Pf}(A) \cdot e_1 \wedge \dots \wedge e_n.$

“Buried” non-density points in the fat Cantor set

Posted on December 1, 2024 by The author

A point $p$ of a Lebesgue measurable set $A \subset \mathbb{R}^n$ is called the Lebesgue density point of $A$ if $\lim_{\epsilon \to 0} \frac{\mu(A \cap B(p,\epsilon))}{\mu(B(p,\epsilon))}=1$ where $B(p, \epsilon)$ denotes the open ball of radius $\epsilon$ about $p$ ; in some sense this means that $A$ takes 100% of the space about $p$ . In general, if the limit above exists (regardless of whether it’s equal to 1 or not), we call it’s value the Lebesgue density of $A$ at $p$ , denoted by $\mathrm{dens}_A(p)$ . Fixing any Lebesgue measurable set $A$ , there’s a theorem called Lebesgue density theorem which states that for almost every $x \in \mathbb{R}$ we have either $\mathrm{dens}_A(x)=1$ or $\mathrm{dens}_A(x)=0$ . The theorem looks reasonable if we think of simple shapes in space, for example if $A$ is a closed filled square in $\mathbb{R}^2$ , then the interior points have density $1$ while the points in the complement have density 0, and the points on the boundary have measure density $\frac{1}{2}$ , but the boundary of the square is a measure zero in $\mathbb{R}^2$ . Restricting our attention to $\mathbb{R}$ and thinking about it for the favorite set in real analysis, Cantor set, it’ll still make sense, because the Cantor set after all has measure zero itself, so those fractions are zero for all of the radii $\epsilon>0$ and point $x \in \mathbb{R}$ . Now one might wonder to twist the situation more by thinking about the fat Cantor set $C$ ; where the set now has some mass in contrast with the Cantor set. Then how do the density and non-density points (i.e. points where that limit is either not equal to 1 or it doesn’t exist) of $C$ look like? The fat Cantor set contains some easy to find Lebesgue non-density points: the boundary points of the open intervals that we remove at each step have density $\frac{1}{2}$ , let such points in $C$ be denoted by $\mathrm{Ends}(C)$ .

Question: Is there any point in $C \setminus \mathrm{Ends}(C)$ that is a non-density point of $C$ ?

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