Due Thursday, February 1 at 1:00 pm.

The second pdf contains some hints for problem 3.

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January 31, 2007 at 6:59 pm |

I believe there to be a typographical error in the definition of the \otimes product for 1d. Specifically, x_4y_3 term should carry a negative sign.

February 1, 2007 at 4:36 am |

Also, maybe this is something way basic for everyone versed in topology, but it would be nice for me if you could clarify that the disks are open. I eventually just said, in part (e), “we must have C be an open disk; this does not change the area, so all important properties are preserved, but the openness will be required later.”

Now to prove… 1e, 2, and 3. Woohoo…

February 1, 2007 at 7:52 am |

Jed,

I think you may be right. Writing a vector (a,b,c,d) as

a + bi + cj + dk the multiplication table should be

i^2=j^2=k^2= -1

ij=-ji=k

jk=-kj=i

ki=-ik=j

*Sigh*. Sorry about this; I should have just written this table and distributivity instead of the huge identity I wrote.

February 1, 2007 at 7:56 am |

Domenic,

Yes, the disk in part (g) -not (e)- is supposed to be open. Although this really doesn’t matter.

February 1, 2007 at 12:11 pm |

No idea what the bracket notation in the hint is supposed to mean. Floor? Ceiling? Round to nearest? Stuff in to a matrix and then take the determinant? Yeah… not much of an issue since I can’t seem to prove the identity anyway (50 more minutes to go!), but yeah.

February 1, 2007 at 12:18 pm |

The bracket notation is the integer part (floor) function. It is the same notation used in lecture. (Edited.)

February 1, 2007 at 12:24 pm |

I must be doing something wrong then, since I have isolated additive terms u and z in the expansion. The fact that there are extra terms besides just u and z means that the ceiling must be > u + z, since any extra contribution besides u and z will push it up by at least 1.