If $V$ is a vector space with inner product and $vinLambda^p V$ and $winLambda^q V$, then $|vwedge...
Suppose we give $V$ an orthonormal basis $e_1,ldots,e_n$, and let $v=sum_I v_I e_I $ and $w=sum_J w_J e_J$, then
begin{align}
|vwedge w|^2& =left| sum_{I,J}v_I w_J e_Iwedge e_Jright |^2
=left| sum_K sum_{K=Isqcup J} (pm1)v_I w_J e_Kright |^2 \
&le sum_K left| sum_{K=Isqcup J} |v_I w_J| right|^2 \
&le sum_Kleft( sum_{Isubset K} |v_I|^2right)
left( sum_{Jsubset K} |w_J|^2right)
end{align}
which is unfortunately more than $|v|^2|w|^2$. Is there any way the statement can be salvaged?
linear-algebra
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
add a comment |
Suppose we give $V$ an orthonormal basis $e_1,ldots,e_n$, and let $v=sum_I v_I e_I $ and $w=sum_J w_J e_J$, then
begin{align}
|vwedge w|^2& =left| sum_{I,J}v_I w_J e_Iwedge e_Jright |^2
=left| sum_K sum_{K=Isqcup J} (pm1)v_I w_J e_Kright |^2 \
&le sum_K left| sum_{K=Isqcup J} |v_I w_J| right|^2 \
&le sum_Kleft( sum_{Isubset K} |v_I|^2right)
left( sum_{Jsubset K} |w_J|^2right)
end{align}
which is unfortunately more than $|v|^2|w|^2$. Is there any way the statement can be salvaged?
linear-algebra
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39
add a comment |
Suppose we give $V$ an orthonormal basis $e_1,ldots,e_n$, and let $v=sum_I v_I e_I $ and $w=sum_J w_J e_J$, then
begin{align}
|vwedge w|^2& =left| sum_{I,J}v_I w_J e_Iwedge e_Jright |^2
=left| sum_K sum_{K=Isqcup J} (pm1)v_I w_J e_Kright |^2 \
&le sum_K left| sum_{K=Isqcup J} |v_I w_J| right|^2 \
&le sum_Kleft( sum_{Isubset K} |v_I|^2right)
left( sum_{Jsubset K} |w_J|^2right)
end{align}
which is unfortunately more than $|v|^2|w|^2$. Is there any way the statement can be salvaged?
linear-algebra
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
Suppose we give $V$ an orthonormal basis $e_1,ldots,e_n$, and let $v=sum_I v_I e_I $ and $w=sum_J w_J e_J$, then
begin{align}
|vwedge w|^2& =left| sum_{I,J}v_I w_J e_Iwedge e_Jright |^2
=left| sum_K sum_{K=Isqcup J} (pm1)v_I w_J e_Kright |^2 \
&le sum_K left| sum_{K=Isqcup J} |v_I w_J| right|^2 \
&le sum_Kleft( sum_{Isubset K} |v_I|^2right)
left( sum_{Jsubset K} |w_J|^2right)
end{align}
which is unfortunately more than $|v|^2|w|^2$. Is there any way the statement can be salvaged?
linear-algebra
linear-algebra
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
edited Nov 28 '18 at 16:25
ancientmathematician
4,4581413
4,4581413
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
asked Nov 28 '18 at 16:04
Monstrous Moonshine
2,6941630
2,6941630
I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39
add a comment |
I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39
I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39
add a comment |
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I think the notation fails to make clear exactly what you are summing over; the outer sum is over $p+q$ sets $K$, but the inner one is over $p$-sets $I$ and by defintion $J=Ksetminus I$ - as written it looks as though both sums are over $K$.
– ancientmathematician
Nov 28 '18 at 16:37
I don't get the first inequality: it seems to me to say that the square of a sum is less than the sum of squares.
– ancientmathematician
Nov 28 '18 at 16:39