David Corfield
sandbox

α X,Y,Z rev=α Z,Y,X 1\alpha^{rev}_{X,Y,Z} = \alpha^{-1}_{Z,Y,X}

Ganesalingam in The Language of Mathematics uses Discourse Representation Theory (DRT) where Ranta uses Dependent Type Theory (DTT). My money is still on DTT. Does anyone using DRT in computerised mathematics?

According to Ganesalingam

DRT was developed to handle Geach’s ‘donkey sentences’ (Geach, 1980) and, as we will show in §3.3.1, there are similar sentences in real mathematical texts. We will also be able to adapt DRT to overcome deficiencies noted by Ranta in his own work; see §3.3.2 for details. (p. 15, n6)

DTT claims to do well with donkey sentences too, as I said in the post.

As far as I can see from the preview, one supposed deficiency of DTT is treated in §3.3.6:

Our approach also correctly handles an area which Ranta highlights as a deficiency of his own analysis. He notes that the plural pronoun ‘they’ is sometimes given a distributive reading and sometimes a collective reading; cf.

Nor do we quite understand the use of the plural pronoun They, which is sometimes distributive, paraphrasable by the term conjunction e.g.

If AA and BB do not lie outside the line aa, they are incident on it

but sometimes used on the place of the “surface term conjunction”, so that it fuses together the arguments of the predicate, e.g.

If aa and bb do not converge, they are parallel.

(Ranta, 1994, p. 12)

Our analysis correctly distinguishes distributive and collective readings. ‘they’ is always taken to be a pronoun introducing an underlined plural discourse referent, requiring a plural antecedent. The difference in the readings of the two sentences given above is accounted for by the fact that ‘incident on it’ is distributive (as a single point is incident on a single line) and the adjective ‘parallel’ is collective (a single line cannot be parallel, but a collection of lines can). (p. 64)

How very odd. I wonder why Ranta believed this to be difficult for DTT? Suppose we have x,y:AP(x),Q(x,y):Propx, y: A \vdash P(x), Q(x, y): Prop and suppose we have a,b:Aa, b: A and P(a),P(b),Q(a,b)P(a), P(b), Q(a, b) all true. Then we would say ‘aa and bb are AAs. They are PP. They are QQ.’ if sugared versions allow this, such as when QQ says of two lines that they’re parallel.

Maybe it’s not clear when such a Q(a,b)Q(a, b) affords the sugaring, ‘They are QQ’. We wouldn’t do this for an asymmetric relation, such as less than. Certainly many symmetric relations are fine, ‘They are parallel/equal/isometric/homeomorphic’.

Mod()Mod(-) and Set ()Set^{(-)}

G(f 1+f 2)= Gf 1+ Gf 2 \int_G (f_1 + f_2) = \int_G f_1 + \int_G f_2
f 1,f 2[G,] Top\forall f_1, f_2 \in [G, \mathbb{C}]_{\text{Top}}

( 3) cpt( 1) +( 3) +( 1) cpt(\mathbb{R}^3)^{cpt} \wedge (\mathbb{R}^1)_+ \union (\mathbb{R}^3)_+ \wedge (\mathbb{R}^1)^{cpt} different from ( 1) cpt( 3) +( 1) +( 3) cpt(\mathbb{R}^1)^{cpt} \wedge (\mathbb{R}^3)_+ \union (\mathbb{R}^1)_+ \wedge (\mathbb{R}^3)^{cpt}

E^ dRE^ (pb) Π(E^) ch E Π dRE^ \array{ \hat E &\stackrel{}{\longrightarrow}& \flat_{dR} \hat E \\ \downarrow &{}^{(pb)}& \downarrow \\ \Pi(\hat E) &\stackrel{ch_E}{\longrightarrow}& \Pi \flat_{dR} \hat E }
E^ dRE^ (pb) <(E^) < dRE^ \array{ \hat E &\stackrel{}{\longrightarrow}& \subset_{dR} \hat E \\ \downarrow &{}^{(pb)}& \downarrow \\ \lt(\hat E) &\stackrel{}{\longrightarrow}& \lt \subset_{dR} \hat E }

/!!/

///\!\!/

T X p 1 X p 2 (pb) i X X i X X \array{ T^\infty X &\stackrel{p_1}{\longrightarrow}& X \\ \downarrow^{\mathrlap{p_2}} &(pb)& \downarrow^{\mathrlap{i_X}} \\ X &\stackrel{i_X}{\longrightarrow}& \Im X }

A *BA_\ast B, ABA \star B, A*BA * B

If (C 2C 1C 0)=C 2C 0C 0\Im(C_2\to C_1\to C_0) = C_2 \to C_0\to C_0, then base change and adjoints.

(A 2A 1A 0)(C 2C 0C 0)(A 2A 1× C 0C 1A 0)(C 2C 1C 0)(A_2 \to A_1 \to A_0) \to (C_2 \to C_0\to C_0) \mapsto (A_2 \to A_1 \times_{C_0} C_1 \to A_0) \to (C_2\to C_1\to C_0)

B 1A 1× C 0C 1 A 1 C 1 C 0 \array{ B_1 \leftarrow A_1 \times_{C_0} C_1 & \to & A_1\\ \downarrow & & \downarrow \\ C_1 & \to & C_0 }

So right adjoint to base change:

(B 2B 1B 0)(C 2C 1C 0)(B 2??B 0)(C 2C 0C 0)(B_2 \to B_1 \to B_0) \to (C_2 \to C_1\to C_0) \mapsto (B_2 \to ?? \to B_0) \to (C_2\to C_0\to C_0)

Is it just to make Hom H/C 1(A 1× C 0C 1,B 1)Hom_{H/C_1}(A_1 \times_{C_0} C_1, B_1) be equivalent to Hom H/C 0(A 1,??)Hom_{H/C_0}(A_1, ??)?

\displaystyle \int, \textstyle \int, \textsize \int, \scriptsize \int, \scriptscriptsize \int

probability, Probability?

Dynkin diagram?/
Dynkin quiver?
Platonic solid?finite subgroups of SO(3)?finite subgroups of SU(2)?simple Lie group?
A n1A_{n \geq 1}cyclic group?
n+1\mathbb{Z}_{n+1}
cyclic group?
n+1\mathbb{Z}_{n+1}
special unitary group?
SU(n+1)SU(n+1)
D4?Klein four-group?
D 4 2× 2D_4 \simeq \mathbb{Z}_2 \times \mathbb{Z}_2
quaternion group?
2D 42 D_4 \simeq Q8?
SO(8)?
D n4{\bgcolor{red} D_{n \geq 4}}dihedron?,
hosohedron?
dihedral group?
D 2(n2)D_{2(n-2)}
binary dihedral group?
2D 2(n2)2 D_{2(n-2)}
special orthogonal group?
SO(2n)SO(2n)
E 6E_6tetrahedron?tetrahedral group?
TT
binary tetrahedral group?
2T2T
E6?
E 7E_7cube?,
octahedron?
octahedral group?
OO
binary octahedral group?
2O2O
E7?
E 8E_8dodecahedron?,
icosahedron?
icosahedral group?
II
binary icosahedral group?
{\bgcolor{red}2I2I}
E8?

\begin{table} \begin{tabular}{l | a | b | a | b} \hline \rowcolor{LightCyan} \mc{1}{} & \mc{1}{x} & \mc{1}{y} & \mc{1}{w} & \mc{1}{z} \ \hline variable 1 & a & b & c & d \ variable 2 & a & b & c & d \ \hline \end{tabular} \end{table}

{ b \color{red} c \color d } e

abcde{\bgcolor{red} a b} c {\bgcolor{#0F0}d e}

a{\bgcolor{red} a}

b{\color{green} b}

c{\bgcolor{green} \color{red} c}

Last revised on July 30, 2020 at 04:47:42. See the history of this page for a list of all contributions to it.