Hodge dual: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>

: This revision was by author [[User:clumma|clumma]] and made on <tt>2016-05-18 15:19:24 UTC</tt>.<br>

: This revision was by author [[User:clumma|clumma]] and made on <tt>2016-05-18 22:03:31 UTC</tt>.<br>

: The original revision id was <tt>~~583498319~~</tt>.<br>

: The original revision id was <tt>583525823</tt>.<br>

: The revision comment was: <tt></tt><br>

The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>

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=Computing the dual=

~~We can see how to go about computing the dual by looking at the dual~~ of ~~wedge products of basis elements. If we have~~ n ~~basis elements B = {b1, b2~~, ~~..., bn}, then if~~ we have a k-~~fold wedge product of k of these basis~~ elements ~~in order, we may subtract the set~~ of ~~these~~ k ~~elements {bi}~~, ~~in the sense~~ of ~~set theory subtraction, from B. In other words, if we have a k-fold product of basis elements X in order, we may take the corresponding~~ (n-k)-~~fold product Y~~ of ~~all~~ the ~~remaining~~ basis elements ~~in order~~. ~~Then X∧Y will be 1 if X concatenated with Y gives an even permutation~~, and ~~-1 if it gives on odd permutation~~. ~~Hence,~~ we ~~may take~~ the ~~(n-i)th element~~ of the k-~~vector, and this becomes the ith element~~ of the ~~dual if~~ the ~~permutation of~~ the ~~k basis elements~~ in ~~order, concatenated~~ with ~~the remaining~~ (n-k) ~~elements in order, is an even permutation~~. ~~If it is an odd permutation, then minus~~ the (~~n-i~~)~~th element becomes~~ the ~~ith element~~ of the ~~dual~~.

Again with a basis of dimension n, suppose we have a k-multival V and wish to find its dual multimonzo M. The elements of V are associated with k-combinations, and of M with (n-k)-combinations, of the basis elements. Because of the symmetry of binomial coefficients, V and M will have the same length. To find M we adjust the signs of V with the following procedure

1. Let C be the k-combinations of the numbers 1..n in lexicographic order

2. C will have the same length as V and M

3. Sum the numbers in each combination Ci with ceil(k/2) to find Si

4. Multiply the ith element of V by -1^(Si)

and then reverse the elements of V.

To find an unknown V from a known M, first reverse M and then adjust the signs.

=Using the dual=

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<br />

<h1 id="toc2"><a name="Computing the dual"></a>Computing the dual</h1>

~~We can see how to go about computing the dual by looking at the dual~~ of ~~wedge products of basis elements. If we have~~ n ~~basis elements B = {b1, b2~~, ~~..., bn}, then if~~ we have a k-~~fold wedge product of k of these basis~~ elements ~~in order, we may subtract the set~~ of ~~these~~ k ~~elements {bi}~~, ~~in the sense~~ of ~~set theory subtraction, from B. In other words, if we have a k-fold product of basis elements X in order, we may take the corresponding~~ (n-k)-~~fold product Y~~ of ~~all~~ the ~~remaining~~ basis elements ~~in order~~. ~~Then X∧Y will be 1 if X concatenated with Y gives an even permutation~~, and ~~-1 if it gives on odd permutation~~. ~~Hence,~~ we ~~may take~~ the ~~(n-i)th element~~ of the k-~~vector, and this becomes the ith element~~ of the ~~dual if~~ the ~~permutation of~~ the ~~k basis elements~~ in ~~order, concatenated~~ with ~~the remaining~~ (n-k) ~~elements in order, is an even permutation~~. ~~If it is an odd permutation, then minus~~ the (~~n-i~~)~~th element becomes~~ the ~~ith element~~ of the ~~dual~~.<br />

Again with a basis of dimension n, suppose we have a k-multival V and wish to find its dual multimonzo M. The elements of V are associated with k-combinations, and of M with (n-k)-combinations, of the basis elements. Because of the symmetry of binomial coefficients, V and M will have the same length. To find M we adjust the signs of V with the following procedure<br />

<br />

1. Let C be the k-combinations of the numbers 1..n in lexicographic order<br />

2. C will have the same length as V and M<br />

3. Sum the numbers in each combination Ci with ceil(k/2) to find Si<br />

4. Multiply the ith element of V by -1^(Si)<br />

<br />

and then reverse the elements of V.<br />

<br />

To find an unknown V from a known M, first reverse M and then adjust the signs.<br />

<br />

<h1 id="toc3"><a name="Using the dual"></a>Using the dual</h1>

The dual allows one to find the wedgie, which is a normalized multival, by wedging together monzos and then taking the dual. For instance from M = |0 3 -2 0&gt;∧|-2 1 -1 1&gt;, which is ||6 -4 0 -1 3 -2&gt;&gt;, considered above, we may find the dual Mº as ||6 -4 0 -1 3 -2&gt;&gt;º = &lt;&lt;-2 -3 -1 0 4 6||. Normalizing this to a wedgie gives &lt;&lt;2 3 1 0 -4 -6||, the wedgie for bug temperament. Then if W is the wedgie for ennealimmal considered above, W∧Mº = &lt;W|M&gt; = 1. We can also take a multival, and use the dual to get a corresponding mulitmonzo, and then use the same method described on the <a class="wiki_link" href="/abstract%20regular%20temperament">abstract regular temperament</a> page for extracting a normal val list from a multival to get a normal comma list from the multimonzo.</body></html></pre></div>

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 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:clumma|clumma]] and made on <tt>2016-05-18 15:19:24 UTC</tt>.<br>
+: This revision was by author [[User:clumma|clumma]] and made on <tt>2016-05-18 22:03:31 UTC</tt>.<br>
-: The original revision id was <tt>583498319</tt>.<br>
+: The original revision id was <tt>583525823</tt>.<br>
 : The revision comment was: <tt></tt><br>
 The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
@@ Line 15: / Line 15: @@
 =Computing the dual=
-We can see how to go about computing the dual by looking at the dual of wedge products of basis elements. If we have n basis elements B = {b1, b2, ..., bn}, then if we have a k-fold wedge product of k of these basis elements in order, we may subtract the set of these k elements {bi}, in the sense of set theory subtraction, from B. In other words, if we have a k-fold product of basis elements X in order, we may take the corresponding (n-k)-fold product Y of all the remaining basis elements in order. Then X∧Y will be 1 if X concatenated with Y gives an even permutation, and -1 if it gives on odd permutation. Hence, we may take the (n-i)th element of the k-vector, and this becomes the ith element of the dual if the permutation of the k basis elements in order, concatenated with the remaining (n-k) elements in order, is an even permutation. If it is an odd permutation, then minus the (n-i)th element becomes the ith element of the dual.
+Again with a basis of dimension n, suppose we have a k-multival V and wish to find its dual multimonzo M. The elements of V are associated with k-combinations, and of M with (n-k)-combinations, of the basis elements. Because of the symmetry of binomial coefficients, V and M will have the same length. To find M we adjust the signs of V with the following procedure
+. Let C be the k-combinations of the numbers 1..n in lexicographic order
+. C will have the same length as V and M
+. Sum the numbers in each combination Ci with ceil(k/2) to find Si
+. Multiply the ith element of V by -1^(Si)
+and then reverse the elements of V.
+To find an unknown V from a known M, first reverse M and then adjust the signs.
 =Using the dual=
@@ Line 29: / Line 38: @@
 &lt;br /&gt;
 &lt;!-- ws:start:WikiTextHeadingRule:4:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc2"&gt;&lt;a name="Computing the dual"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:4 --&gt;Computing the dual&lt;/h1&gt;
-We can see how to go about computing the dual by looking at the dual of wedge products of basis elements. If we have n basis elements B = {b1, b2, ..., bn}, then if we have a k-fold wedge product of k of these basis elements in order, we may subtract the set of these k elements {bi}, in the sense of set theory subtraction, from B. In other words, if we have a k-fold product of basis elements X in order, we may take the corresponding (n-k)-fold product Y of all the remaining basis elements in order. Then X∧Y will be 1 if X concatenated with Y gives an even permutation, and -1 if it gives on odd permutation. Hence, we may take the (n-i)th element of the k-vector, and this becomes the ith element of the dual if the permutation of the k basis elements in order, concatenated with the remaining (n-k) elements in order, is an even permutation. If it is an odd permutation, then minus the (n-i)th element becomes the ith element of the dual.&lt;br /&gt;
+Again with a basis of dimension n, suppose we have a k-multival V and wish to find its dual multimonzo M. The elements of V are associated with k-combinations, and of M with (n-k)-combinations, of the basis elements. Because of the symmetry of binomial coefficients, V and M will have the same length. To find M we adjust the signs of V with the following procedure&lt;br /&gt;
+&lt;br /&gt;
+. Let C be the k-combinations of the numbers 1..n in lexicographic order&lt;br /&gt;
+. C will have the same length as V and M&lt;br /&gt;
+. Sum the numbers in each combination Ci with ceil(k/2) to find Si&lt;br /&gt;
+. Multiply the ith element of V by -1^(Si)&lt;br /&gt;
+&lt;br /&gt;
+and then reverse the elements of V.&lt;br /&gt;
+&lt;br /&gt;
+To find an unknown V from a known M, first reverse M and then adjust the signs.&lt;br /&gt;
 &lt;br /&gt;
 &lt;!-- ws:start:WikiTextHeadingRule:6:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc3"&gt;&lt;a name="Using the dual"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:6 --&gt;Using the dual&lt;/h1&gt;
 The dual allows one to find the wedgie, which is a normalized multival, by wedging together monzos and then taking the dual. For instance from M = |0 3 -2 0&amp;gt;∧|-2 1 -1 1&amp;gt;, which is ||6 -4 0 -1 3 -2&amp;gt;&amp;gt;, considered above, we may find the dual Mº as ||6 -4 0 -1 3 -2&amp;gt;&amp;gt;º = &amp;lt;&amp;lt;-2 -3 -1 0 4 6||. Normalizing this to a wedgie gives &amp;lt;&amp;lt;2 3 1 0 -4 -6||, the wedgie for bug temperament. Then if W is the wedgie for ennealimmal considered above, W∧Mº = &amp;lt;W|M&amp;gt; = 1. We can also take a multival, and use the dual to get a corresponding mulitmonzo, and then use the same method described on the &lt;a class="wiki_link" href="/abstract%20regular%20temperament"&gt;abstract regular temperament&lt;/a&gt; page for extracting a normal val list from a multival to get a normal comma list from the multimonzo.&lt;/body&gt;&lt;/html&gt;</pre></div>