1

where $ \mathbf u $ and $ \mathbf v $ are vectors. From Linear and Geometric Algebra by Alan Macdonald.

user997712
  • 195
  • 1
  • 9
  • I think there should be + instead of - in RHS – iostream007 Jun 23 '13 at 10:27
  • The print definitely has a '-'. How would you prove it if it were a '+'? – user997712 Jun 23 '13 at 10:37
  • What is a multivector?If the $\wedge$ is the cross product of vectors, then I would say that $u$, $v$ are in $\mathbb R^3$. Am I right? – Avitus Jun 23 '13 at 10:53
  • Are you sure $u$ and $v$ are multivectors instead of just vectors? Chris's answer doesn't seem to work if they're actually multivectors. – Muphrid Jun 23 '13 at 18:55
  • @Murphid - I edited the question. It was a false assumption on my part. – user997712 Jun 23 '13 at 20:26
  • Your sign is obviously wrong. To see this, just take any non-zero $\mathbf u, \mathbf v$ with $\mathbf{u.v} = 0$. – TonyK Jun 23 '13 at 20:36
  • @TonyK Taking $ u = e_1 , v = e_2 $ I get L.H.S = 1, R.H.S $ = 0 - (e_1\wedge e_2)^2 = 1 $. Where is the problem? – user997712 Jun 23 '13 at 20:41
  • @user99712: If $\mathbf u$ and $\mathbf v$ are ordinary vectors, then $\mathbf{(u \wedge v)}^2$ is positive, isn't it? – TonyK Jun 23 '13 at 20:45
  • Oh, I see. $\mathbf u \wedge \mathbf v$ is to be interpreted as a multivector. But you edited your question to remove all references to multivectors, so I didn't pick up on that... – TonyK Jun 23 '13 at 20:49
  • In the context of this being an expression in geometric algebra (a clifford algebra over the real numbers), it should be clear that this is a wedge product (and not a cross product just confusingly denoted with wedge). Still, a couple people have had this issue. What can we do (with tags, or something else) to make the proper context clearer? – Muphrid Jun 23 '13 at 20:51
  • @Muphrid: Tag it as "not for the over-50's". I never learnt this stuff at university... – TonyK Jun 23 '13 at 21:07
  • Yeah, while Hestenes was working on this stuff in the 70s, I don't think it's really taken off except in the last 20 years. And by taken off, I mean "worked on by more than a few enthusiasts". But I think anyone with enough background in vector algebra can come to appreciate the advantages of geometric algebra over index notation and differential forms. – Muphrid Jun 23 '13 at 21:20
  • @TonyK: Neither did I, so I am learning now. My main goal is to be able to learn physics with GA, since it has been called a unifying language for physics. – user997712 Jun 23 '13 at 21:25

3 Answers3

1

In a paper i read $$a\cdot b=\dfrac {ab+ba}{2}=b\cdot a$$

and $$a\wedge b=\dfrac {ab-ba}{2}$$ $$(u\cdot v)^2-(u\wedge v)^2$$ $$\left(\dfrac {uv+vu}{2}\right)^2-\left(\dfrac {uv-vu}{2}\right)^2$$ $$\left(\dfrac {uv+vu}{2}-\dfrac {uv-vu}{2}\right)\left(\dfrac {uv+vu}{2}+\dfrac {uv-vu}{2}\right)$$ $$(vu)(uv)$$ $$(uv)^2$$

iostream007
  • 4,529
  • I'm not sure it is fully correct but i do this solution on basis of linked paper – iostream007 Jun 23 '13 at 11:07
  • @ionstream007 are $a$ and $b$ vectors? Then what is $ab$, $ba$? – Avitus Jun 23 '13 at 11:20
  • @Avitus $a$ and $b$ are multivectors, i.e. elements of the Clifford algebra of some vector space. See here: http://en.wikipedia.org/wiki/Geometric_algebra – Chris Taylor Jun 23 '13 at 11:23
  • ab is scaler while a and b are multivectors.I've attached a paper in solution – iostream007 Jun 23 '13 at 11:24
  • I don't understand the last step. Actually you can just start from the R.H.S, $$ (u\cdot v + u\wedge v)(u\cdot v - u \wedge v) = (u\cdot v + u\wedge v)(u\cdot v + v \wedge u) = uvvu $$ – user997712 Jun 23 '13 at 11:28
  • @iostream007 $uv$ is a not a scalar quantity in general. For example consider the geometric algebra $Cl(\mathbb{R}^2)$ with $u=u_1e_1+u_2e_2$ and $v=v_1e_1+v_2e_2$. Then $uv = u_1v_1 + u_2v_2 + (u_1v_2-u_2v_1)e_1e_2$. – Chris Taylor Jun 23 '13 at 11:32
  • No offence to anyone.I don't know about multivectors (I know vector calculas that used in physics).In the paper I've read that $a^2=|a|^2\ge 0,$ a is a positive scaler. and it is also written that scaler means real number so is there any difference in uv and vu. – iostream007 Jun 23 '13 at 11:46
  • @iostream007 No offence is taken. It's true that when $a$ is a multivector, $a^2$ is a positive scalar. But multiplying two multivectors $u$ and $v$ does not give a scalar in general. Since $uv=u\cdot v + u\wedge v$, you have $vu = v\cdot u - v\wedge u = u\cdot v - u\wedge v$ so there is no simple relationship between $uv$ and $vu$ in general (unless $u\cdot v=0$ or $u\wedge v=0$). – Chris Taylor Jun 23 '13 at 12:33
  • @ChrisTaylor so if you can correct my answer,your welcome,cause I've not any more thing to say.And it is also new field for me thanks – iostream007 Jun 23 '13 at 13:18
  • It's not true that $a^2$ is a scalar for an arbitrary multivector; a good example is the Faraday bivector in special relativity. – Muphrid Jun 23 '13 at 18:56
1

This is straightforward, then, using grade projection and associativity of the geometric product. Consider the product $vuuv$:

$$vuuv = u^2 v^2 = \langle vuuv \rangle_0$$

On the other hand, you can group the products like so

$$vuuv = (vu)(uv) = (v \cdot u) (u \cdot v) + (v \wedge u)(u \wedge v) = (u \cdot v)^2 - (u \wedge v)^2$$

Because you know from the first equation that $vuuv$ is a scalar, this is all you need to consider--for instance, there are some bivector terms that I didn't write down, but you can argue that the overall bivector component must be zero, and thus you don't even need to compute them.

Edit: this basic technique of using grade projection and associativity is very useful. You can, for instance, prove the BAC-CAB rule without using index notation this way (but you do have to approach it with some cyclic products, which is kinda lame). It's also very useful when you start doing geometric calculus to prove some basic vector calculus identities that would otherwise be cumbersome.

Muphrid
  • 19,902
  • Thanks for the insight. Although i am not sure i have encountered the BAC-CAB rule yet. – user997712 Jun 23 '13 at 20:43
  • It's the identity $a \cdot (b \wedge c) = c( a \cdot b) - b (a \cdot c)$. When written with cross products there's an overall sign change, which is why it's called the BAC-CAB rule. – Muphrid Jun 23 '13 at 20:45
0

You can use

$$ \begin{align} (u\cdot v)^2 - (u\wedge v)^2 & = \frac{1}{4} \left( (uv+vu)^2 - (uv-vu)^2\right) \\ & = \frac{1}{4} \left(uvuv + uvvu + vuuv + vuvu - uvuv + uvvu + vuuv - vuvu \right) \\ & = \frac{1}{2} \left( uvvu + vuuv\right) \\ & = \frac{1}{2} \left( u^2v^2+u^2v^2\right) \\ & = u^2v^2 \end{align} $$

The third line to the fourth line is true because

$$uvvu = uv^2u = uuv^2 = u^2v^2$$

where we can re-order terms because $u^2$, $v^2$ are scalars.

Chris Taylor
  • 28,955