# Projective Geometry: Why is multiplication defined this way?

I am trying to understand this new way of multiplying in projective geometry.

Why is it defined like this? Also does this have anything to do with multiplication using a slide ruler? (The picture in the link shows that $4 \cdot 4 = 16$ and $4 \cdot 2 =8$. Every unit is a power of 2. Slide rulers were commonly used in the old days way before the use of a calculator.)

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Not sure if the your second picture is relevant in this context...Is there a reason why you would want us to show a slide rule? –  user21436 Apr 20 '12 at 2:55
No, slide rules use logarithms. The multiplication above just relies on ratios being the same. –  copper.hat Apr 20 '12 at 3:07
This isn't a complete answer, but worth considering: The reason we use a product with four inputs (cross-ratio) is that there's no meaningful product with just 3 respected by linear fractional transformations. Any three points in projective space can be mapped to any other three via a linear fractional transformation, so we need at least 4 inputs to define any kind of measurement that doesn't depend on your model of projective space. –  Brett Frankel Apr 20 '12 at 3:15
Let me clarify what I'm seeing here: Look at the first diagram. Let a = 2 and b = 3 then ab = 6. Align 1 from the top so that it matches with b at the bottom and align a from the top so that it matches with ab at the bottom. This is the same as using the slide ruler that I've drawn. –  Low Scores Apr 20 '12 at 3:36
Not that it matters, but I used slide rules in school. The slide rule just adds the logs of numbers. The geometry method actually produces a length that is the product of $a$ and $b$, whereas the slide rule produces a length that is the sum of $\log a$ and $\log b$. So, you need a lot more room to multiple with the geometric method. –  copper.hat Apr 20 '12 at 4:10

# Error in image

First off, your image looks wrong to me. In my opinion, you'd want the following elements (too lazy to draw an image just now):

• a line $g$ connecting $0$, $1$, $a$ and $b$
• a different line $h$ through $0$
• a point $p$ on $h$, perhaps directly above $1$
• a triangle $t$ connecting $1$, $a$ and $p$
• a similar triangle $b$, $a\cdot b$ and $p'$

In other words, the construction would be:

1. choose $p$ arbitrarily
2. $p'$ is the point where (the connection of $0$ and $p$) intersects (the line parallel to $1\vee p$ through $b$)
3. $a\cdot b$ is the point where (the parallel to $a\vee p$ through $p'$) intersects (the line $0\vee 1$)

In that case, your image would represent the Euclidean version of multiplication, which is a special case of the projective version. In your image, the triangle over $1$ and $b$ looks similar to the one over $b$ and $a\cdot b$, so it appears that you'd have constructed $b^2$ instead of $a\cdot b$.

http://www-m10.ma.tum.de/bin/view/MatheVital/GeoCal/GeoCal4x2a has applets of both the Euclidean and the projective view of these situations, although the text is in German.

# First question

For your first question, I guess the simplest answer would be “because it works”. So what is the objective? Given a projective scale along a line, i.e. the points $0$, $1$, $a$, $b$ and $\infty$, one wants to construct the point $a\cdot b$ using only tools from projective incidence geometry, i.e. joins (lines connecting points) and meets (intersection points of lines). There aren't many configurations which can accomplish this, but the von-Staudt construction your immage suggests can accomplish this, as you can check for the euclidean case and generalize as all operations can be interpreted projectively.

# Second question

I see only a slight connection to the slide ruler. Both use some isomorphism to make multiplication accessible in a geometric way. One uses logarithms to translate multiplication into addition, whereas the other transforms it to the algebra of projective geometry. Apart from that, I see no connection.

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