## “now here’s something you’ll *really* like”… “that trick *never* works”

Posted by vlorbik on November 23, 2010

i’m actually kind of reluctant to post this;

it’s probably the best idea i’ve had since

i started making up lectures without words

and now it’ll be easy to steal. you saw it

here first.

the Big triangle is made up of seven Little triangles.

each Little has seven Points. moreover the Points

of each Little (considered as subobjects of their Little)

are arranged in the same pattern

as the Littles themselves are (considered

as subobjects of Big).

pick any Little and call it L1.

there are three dark points in L1.

find the three Littles that correspond

to these three Points.

(for example, let L1 be the lower-left Little;

the three Littles i refer to now run along

the right side of Big [just as the dark Points

run along the right side of L1]).

the three Littles in question are then

precisely those in which the Point…

P1, say… that corresponds to L1

(considered as a subobject of Big)

is dark. i’ll be here all week.

(it’s better with the handwaving.)

## Sue VanHattum said

James Tanton has a whole book of Math Without Words. His are all puzzles, and yours here is one of those theorems without words, I think. I don’t get it yet.

I see how a dark spot on one triangle corresponds to another triangle, which has a dark spot corresponding to the first triangle. mI see that it is allowed for a triangle to refer to itself that way (left and right middle do), and that each one has 3 dark spots.

How does one choose which spots go dark, and what does it have to do with the equation? Does that say P2(Z2) = P2(Z2)*? (I think you said somewhere that the star means dual, so I’m guessing it’s all 2’s, though the last looks like a 1.)

## vlorbik said

right; there’s a “duality” theorem behind all this.

P_2(F_2) is here a set of seven (so-called) points.

sometimes one calls ’em something like

[1], [2], … , [7] but i’ll just spare myself

the editing time and call ’em 1, … ,7.

[actually, any moment now i’m gonna want

the binary representation so i’ll go ahead

right now on second thought and admit that

these are themselves just nicknames for

001, 010, … ,111

(which are themselves “really” just nicknames

for the *official* names:

(0,0,1), (0,1,0), … (1,1,1)…

these objects will have started life

as the seven vectors connecting

the origin of a certain co-ordinate

system to a corner of its “unit cube”.

i’ve got some drawings but the margin

is too small to etcetera.)]

the eye-in-the-pyramid-lookin’ thing…

the “triangle pattern” that appears eight

times here… is then produced by arranging

these Points in such a way that seven Lines

are produced (these correspond to certain

planes through the origin in our “cube” model…

2-dimensional subspaces of F_2 (we’re “working

over” the-field-with-two-elements…

F_2 = Z_2 = { [0]_2 , [1]_2 }

if you just *have* to know [though really

one leaves off

the subscripts in actual calculations…

more “nicknames”…]) in other words.

the Lines of P_2 (associated with certain

planes of (F_2)^3 as i’ve just remarked)

can then be determined algebraically

using the “dot product” of vector analysis.

not done. posting anyway. i’ll be back.

## vlorbik said

the Lines … i *said* i’d be back… are

L_1 = {2, 4, 6}

L_2 = {1, 4, 5}

L_3 = {3, 4, 7}

L_4 = {1, 2, 3}

L_5 = {2, 5, 7}

L_6 = {1, 6, 7}

L_7 = {3, 5, 6}…

but *really* they’re

L_001 = {010, 100, 100}

L_010 = {001, 100, 101}

…

L_111 = {011, 101, 110}.

the point here being that the Point

“abc” (with a, b, c \in {0,1} )

in on the Line L_ABC

(A, B, C \in {0,1}, natch)

precisely when the “dot product”

aA + bB + cC

is equal to zero (as an element

of F_2 of course).

see the previous post for some visuals.

https://vlorblog.wordpress.com/2010/11/23/perhaps-this-will-refresh-your-memory/

the sets of Points and of Lines are the P_2(F_2) and P_2(F_2)*

of the title (of my drawing). abstractly, the “dual space” of

a “space” is the set of functions-to-the-base-field

(that preserve the algebraic structure); the dot-products

in question show that the dual space of the set-of-Points

can usefully be confused with the set-of-Lines (which itself

has the same “shape” as the set-of-Points itself…).

## Sue VanHattum said

I’m wondering if you could start me out with context. This all sounds more interesting when I think about how the Platonic solids pair up into duals (the square has 6 faces and 8 vertices, the octagon has 8 faces and 6 vertices), with the tetrahedron having to act as its own dual.

So F is for field, and P is for what? Point? Plane? Something else?

## vlorbik said

it remains only to choose a pleasing arrangement of {1, 2, … 7}

to label the eye-in-the-pyramid points.

there are, as it’ll turn out, 168 of ’em that’ll work.

and oh, yeah, to remark that one of our “lines” in

such drawings-of-seven-point-projective-space

appears to be a circle. pay this no mind. we’re

lucky to get as good a picture as this. in higher

dimensions the drawings get less symmetrical.

https://vlorblog.wordpress.com/2010/11/23/perhaps-this-will-refresh-your-memory/

## vlorbik said

in P_2(F_2), “P_2” stands for “2-dimensional Projective space”

and “F_2” for “the Field with 2 elements”.

makes this slightly clearer to some eyes.

so our context is “projective spaces”.

there are (on my very limited understanding)

two main ways to go about “constructing”

such spaces mentally.

i’ve sketched an outline of the “top-down” picture.

start with a vector space. , say

(the ordinary 3-D (x, y, z)-space of standard analysis…

“real three-dimensional space” as the saying has it.

now consider the set of

lines through the originin this space. this set… with the “topology” derived

from our initial VS (“vector space”; R^3 in the example)…

is the

projective spaceassociated to our“ordinary” space (remark: it has “dimension” *one less*

than that of the VS it’s associated to… the “lines”

in VS have collapsed to “points” of PS).

back in R^3. consider the unit sphere (S^3):

{ (x, y, z) \in R^3 | x^2 + y^2 + z^2 = 1 }.

any “line through the origin” of R^3 passes

through the sphere *twice*, in a pair of

antipodal points(like the northand south poles [the best-known example]).

so P_2(R_2) is antipodal-pairs-on-the-globe.

some people including me find this easier

to “visualize” than lines-through-the-origin.

in particular, i can more clearly speak of

the “topology”… the question of “what points

are ‘close to’ what points”… since one can

draw little disks on the globe (antipodal

*pairs* of such disks, actually) to illustrate

(–literally!) the situation on a globe.

or on a piece of paper. P_2(R) can

be represented by “identifying points”

along the boundary of a square…

maybe you’ll have seen the “moebius

strip” drawn in this way (identify

a pair of opposite edges with

opposite orientations; on this model

one simply “sews a disc” along the

circle making up the boundary

of the moebius strip to produce

P_2(R)).

but enough of this hilarity.

on the “bottom-up” construction,

one begins with “ordinary” n-space

(over some field) and puts in *extra*

points (“ideal” points usually thought

of as “points at infinity”). more anon.

maybe.

anyhow, part of the fun here is that in projective spaces,

*any* two lines now meet at a point (perhaps “at infinity”),

just as *any* two points of (ordinary *or* projective) space

determine a unique line.

indeed, in P_2, a duality principle holds whereby

any true theorem *remains* true when “point”

is interchanged everywhere with “line” (mutadis

mutandis). amazing stuff really. projective space

turns out in some ways (“keep it simple”) to

be logically *prior* to euclidean space.

## Sue VanHattum said

Hmm, the concept is so wacky at first… I think pictures that get at the conceptual stuff should come before pictures that get at the relationships among elements. (Am I right that the picture above does this?)

## vlorbik said

each point “says” (in effect),

“the three lines-through-me

are

here,here,and

here“. not thatone couldn’t already see that

already (where lines are *easy*

to see… but what about that

pesky *circle*?)…

## vlorbik said

owen by the way.

context?

doggone it…

i should’uh said long ago.

think of the “dot product”.

better still, think of some

finite dimensional vector space V.

i.e., essentially, V = F^n

.

while we’re at it, think of V as

the n-dimensional *column* space

over (the field) F.

thus, V = { [ v_1, v_2, … , v_n]^T : v_i \in F}

(the vector-space V [considered as a set]

consists of all the possible “n-tuples” of

elements from the field F [reading downward]).

typically in lower-division college maths,

F is the Real Number field and n=3;

this is the “ordinary 3-dimensional space”

of the contemporary (age of science)

worldview. x-y-z space; cogito ergo wow.

thingum again!

the so-called “dot product” of two vectors

in this context… which is defined (as we will

see**) as an *algebraic* object

[

which is to say, by applying the

*algebraic* operations built into

the so-called “field axioms”…

namely, additions(-and-subtractions),

and multiplications(-and-divisions)…

to certain “objects” (for instance,

to numbers or variables)

],

the “dot product” of two vectors, i say,

this *algebraically defined* gizmo,

turns out to be the crucial tool in

computing… and finally in talking about…

the *angle* formed by the two vectors

(and several closely-related *geometric*

properties of the situation at hand).

finally… oh sue v… or any other reader…

the relation to “duality” as i keep gesturing at

in blogetty-blogblogg-blogpost after BP

is along these lines.

v \dot w = 0

means (among other things)

that v and w (“vectors”)

are *at a right angle*

to each other.

one consequence of this is that

any line-through-the-origin

determines (“algebraically”!)

a unique plane-through-the-origin

(and vice-versa):

P= {w : w \dot v = 0} kinda thing.

and then you can turn right back around

and get the lines back from the planes.

that’s “duality”.

something very similar… some would

say “the *same* thing”… happens

again and again in situations where

“3-dimensional ‘real’ number space”

is replaced with “n-dimensional space

over an *arbitrary* number-field”:

every true statement about a given

r-dimensional subspace W

in this context will correspond

(“automatically”… which is to say,

because of some theorem-of-duality)

to a true statement about the

“dual space” of W (which is an

(n-r)-dimensional subspace of V).

it’s a “dot product” thing here, too

when you get down to it. which was

sort of my point.

**but. you know what.

i think i promised upthread to *define* that SOB.

but. to hell with it. are there no prisons.

are there no workhouses. should google

go broke just because they hate my browser.

look it up. thank me later.

V.

oh. ps. dammit.

there’s this bit where

n-dimensional *column* space (over F)

and

n-dimensional *row* space (over F)…

sue v and other teachers of “linear algebra”

will have already taken the point…

can be used to *generalize*

the (so-called) dot product.

(that’s why i made such a fuss about

V being a column-space… by habit

i wanted to be sure about the left-right

issues *just in case it came up*.)

the beginning of wisdom

is the study of finite-dimensional

vector spaces.

## suevanhattum said

Reading. Thinking. Show me how this relates to Spot It, and I’ll be quite happy.

## vlorbik said

the Spot It lecture.

let F be the field of seven elements:

F = (S, *, +), where S is some 7-set

({0,1,2,3,4,5,6}, say) and * and +

denote multiplication and addition

(“mod seven”).

F^2 is then a collection of 49 “points”

{(x,y): x \in F, y\in F}; it’s convenient

to think of this as “the (affine) *plane*

over F^2” and picture some such

array as

(0,0) (1,0) (2,0) … (6,0)

(1,0) (1,1)… (6,1)

…

(6,0)… (6,6).

the “lines” in this space are then

solution sets for linear equations

Ax+By=C.

suchlike “lines” can be visualized as

modifications of ordinary R^2 lines.

the relevant modifications are

(1) “throw out points having one or more

non-integer co-ordinates” and

(2)

“use the PacMan topology”

(one will recall, for example, that when

PacMan moves out of the right-hand

edge of his world, he comes back

at the corresponding spot on the

*left*-hand edge [and vice-versa];

likewise for top and bottom.

mod-7 arithmetic on the F^2 array

produces the same result visually

[essentially as a result of 6+1 =0…

this “wraparound” feature, by the way,

is why Rings are called “rings”]).

_ =____

=______

______=

_____=_

____=__

___=___

__=____,

with any luck, will appear as a sketch

of “x+y = 1”, for example.

anyhow, this collection of points-and-lines

is a pretty useful object in its own right…

but it’s not yet what we need for Spot It.

trouble is… while any two *points* determine

a unique *line*, it’s *not* true that any two *lines*

determine a unique *point* (instead, certain pairs

are “parallel” and have no point of intersection).

one can rectify this situation by “adding points”

to the geometry. specifically, one “adds in”

eight so-called “infinite points” (one for each

possible “slope” [the idea that “parallel lines

meet at infinity” is here made formal] and

one more infinite point for the “vertical” lines).

alas, the arithmetic gets slightly complicated.

the most convenient way to supplement our

array turns out to be to consider instead

certain collections-of-points of F^3.

the array considered earlier is, as it were,

“lifted up” into three space and place into

the plane z = 1. thus,

(0,0,1) (1,0,1)…(6,0,1)

(0,1,1) (1,1,1)…(6,1,1)

…

(0,6,1)… (6,6,1).

the “infinite points” are now

{(0,1,0), (1,1,0), (2,1,0)…(6,1,0)}\union

{(1,0,0)}.

drawing a picture might help to see that

these points are a complete set of representatives

of “lines through (0,0,0)” in F^3 (every such line

will contain exactly one of our 57 points).

it’s worthwhile to notice here that none of these

points is (0,0,0) and that the last non-zero

co-ordinate of each is 1. (the formalism is

that of “homogeneous co-ordinates”; this

need not concern us now.)

finally, let G and G* be copies of this space;

write G = { (x, y, z) } and G* = { [a, b, c] }.

define a product, “\dot” (using “infix” notation):

[a,b,c] \dot (x,y,z) = ax+by+cz (mod 7).

(so \dot maps G*\cross G to F).

the lines of P^2(F)… the projective plane over

the field with seven elements are then defined

by fixing an element g \in G* and taking all the

points of G that “dot to zero” with g:

L(g) = {x \in G : g \dot x = 0}.

well, it turns out that this is exactly what we need

to get, not only “two points determine a unique line”,

but also “two lines determine a unique point”.

moreover, each *point* of G* is identified with

a *line* of G… but the structures G and G*

are identified by the *same* set-of-triples

(one has merely used [,] versus (,) for

extra clarity).

now we can make a Spot It deck.

create 57 easily-distinguished cartoons

and put ’em into one-one correspondence

with points of G.

make up 57 cards consisting of the “lines”

in G (as determined by the “dot product”

rule):

~~seven~~eight cartoons on each card,with each card-pair sharing exactly one

cartoon. throw out two cards, alas. done.

@sue v.

feel free to reprint in whole or in part;

i’d’ve run it myself in MMW if i hadn’t’ve

sworn off the prove-you’re-human

timesink frustrationware provided by

google. better still, i suppose, would be

to rewrite it in your own language for

your readers…

## Sue VanHattum said

You tried to show me. I am a little bit happy (because I see there’s lots I’ll get to learn later). I don’t really get it, so can’t put it in my own words yet. I look forward to learning more about all this. Is there a text with lotsa problems you’d recommend? I think this might be a fun project for this summer (and I *will* be done working on my book by then).

## vlorbik said

there’s a more-or-less standard university course

in geometry whose target audience typically consists

largely of future teachers; the thinking is that by considering

*other* axiom systems than the standard (euclidean plane),

students will come to a better understanding of what’s going

on *in* the euclidean context (and, with any luck, stop

thinking of *anything* as “obvious” until it’s been proven

and worked with for a while… this is the dreaded

“course in proving”. math majors should take some

version, too, but seldom do, i think).

anyhow, typically one considers the geometries

typically *called* “non-euclidean” (the parallel postulate

is here replaced either with “no parallels exist” or

with “infinitely many parallels to a line exist through

each point not on the line” [respectively “elliptic” and

“hyperbolic” geometry]) *and* (not “euclidean” but

also, somewhat weirdly, usually not *called* “non-

-euclidean”) others… projective geometries, for example.

the existence of small examples like the fano plane

(the photo of this post is a version) make PG’s rather

a charming subject for those of my bent (start with

the easiest interesting example and try to understand

it better and better). the *graphical* aspect is of course

another charming aspect of geometry (graphically

conceived; one is moving *away* from the pictures

typically [as the concepts sink in]…).

anyhow, such a class is where i saw first saw it;

i sat in (as an unofficial audit) in a version taught

by andrew lenard at IU in the 80’s.

then i taught a similar course in the early 90’s

when i was briefly a salaried pro.

i forget what text i used though.

bruce meserve _fundamental_concepts_of_geometry_

is a cheap dover reprint covering all this ground

masterfully. most instructors would probably reject

it as too mathy… contemporary textbooks typically

take a lot more pages to cover less ground.

i can’t put my hand on my copy just now.

the same could be said of dan pedoe’s

_geometry:_a_comprehensive_course_

(chapters vii & viii… but i haven’t studied

these nearly as carefully as the meserve).

i’ll go out on a limb here and conjecture that

howard eves _a_survey_of_geometry_ (1963)

(volume 1, chapter 6: projective geometry)

is something of a standard source.

that’s all that comes to hand. i’m moving so

my library’s a mess. there may be more in there

somewhere…. but of course the best thing to do

is to poke around in *your* (university’s) library.

check a bunch of indices; look the right stuff

in the right style; voila.

## Sue VanHattum said

Easier to buy the Meserve online than to find parking in Berkeley, make sure my access to their libraries is up-to-date, and get to the math library when it’s open. On my list.

## vlorbik said

but don’t let the rambling fool ya.

the set {(x,y,1)}\union{(x,1,0)}\union{(1,0,0)}

and the “dot product” operation are *all* we need

to generate a Spot It deck (in a “braindead” way…

forget the geometry and just *compute*).

each of our 7^2 + 7 + 1 (=57) objects is “dual”

to a set-of-8 (a “line”); these lines can be

computed with “brute force” (or a computer;

“python” is alleged to be well-suited for these

computations by those skilled in such matters).

i’ll retain the brackets for the dual space…

we can think of these points as “functions”…

but i’ll drop the commas and parens

(typing is hard).

thus, in an easy example, [100] is dual to

the line x=0 (just as in ordinary 3-D *real*

vectorspace): 1x + 0y + 0z = 0 (mod 7)

gives us

L([100]) = {001, 011,021,031,041,051,061, 010}.

sightly harder: consider [111].

now we need x+y+z = 0 (mod 7) so

it’s a matter of considering digit sums:

L([111]) = {061, 151, 241, 331, 421, 511, 601, 610}.

(note that [of course!] these two lines share

a single point: 061 [the first to “spot it” wins]).

for L = L([231]), say?… more fiddling around is required,

but it’s the same game basically. there’s a “finite”

point on this line having x=0. let’s see: 3y+1=0;

3y= -1 = 6; y=2;

021\in L.

now for x=1. 2*1 + 3y +1 =0; 3y = -3; y=-1=6:

161 \in L.

so on. pictures help. it’s sort of relaxing if you can

get in the right groove… very like filling in crosswords.

and the *understanding* will come through the cracks

eventually… with any luck. so my best advice to you

if you want to see what i’ve been going on about

for these two years or so is to replace “7” with “3”

(any prime works) and make yourself a 13-card

Spot-It-deck analogue (with 4 cartoons to a card;

with 1 cartoon common to any card-pair).

no parking at berkeley required.

## vlorbik said

https://vlorbik.wordpress.com/2015/03/28/fanos-rainbow/

colorized at last