MHB Symmetric/Alternating k-linear functions, Wedge Product

[joypav]

I am working through Tu's "An Introduction to Manifolds" and am trying to get an understanding of things with some simple examples. The definitions usually seem simple and understandable, but I want to make sure I can use them for an actual function.

I've worked a few problems below that my professor suggested and just want to check that I am getting it right.

1.
If $f(x,y) = 2x + 3y$, find $Af$ and $Sf$, (the alternating and symmetric k-linear function).
I'm given,
$Sf = \sum_{\sigma \in S_k} \sigma f$
and
$Af = \sum_{\sigma \in S_k} (sgn \sigma) \sigma f$
Solution:
$Sf = \sum_{\sigma \in S_2} \sigma f = (2x+3y)+(2y+3x) = 5x + 5y$
Seems reasonable, as $\sigma f = f$ for any $\sigma$.
$Af = \sum_{\sigma \in S_2} (sgn \sigma) \sigma f = (1)(2x + 3y)+(-1)(2y+3x)=-x+y$
Also seems reasonable, as $\sigma f = (sgn \sigma)f$ for any $\sigma$.

2.
If $g(x,y)=x+y$, show $Ag=0$ and $Sg=g$.
Solution:
$Sg = \sum_{\sigma \in S_2} \sigma g = (x+y)+(y+x)=2x+2y$
Though I am supposed to show that $Sg=g$. Why does it? Of course $g$ is already symmetric, but doesn't the definition for $Sg$ give $2x+2y$? Which is also symmetric.
$Ag = \sum_{\sigma \in S_2} (sgn \sigma) \sigma g = (1)(x+y)+(-1)(y+x)=0$

3.
If $f(x,y)=2x+3y$ and $g(x,y)=4x+5y$, find $f \otimes g$ and $f \wedge g$.
I am given, (for f k-linear and g l-linear)
$( f \otimes g )(v_1, ..., v_{k+l})= f(v_1,..., v_k)\cdot g(v_{k+1},..., v_{k+l})$
and
$f \wedge g = \frac{1}{k!l!}A(f \otimes g)$
Solution:
$( f \otimes g )(x, y, a, b) = (2x+3y)(4a+5b)=8xa+10xb+12ya+15yb$

$f \wedge g = \frac{1}{2!2!}A(8xa+10xb+12ya+15yb)$
I am unsure about both of these. If that's right, then next I'd find the alternating 4-linear function that comes from $8xa+10xb+12ya+15yb$.

 

[GJA]

Hi joypav,

1.
If $f(x,y) = 2x + 3y$, find $Af$ and $Sf$, (the alternating and symmetric k-linear function).
I'm given,
$Sf = \sum_{\sigma \in S_k} \sigma f$
and
$Af = \sum_{\sigma \in S_k} (sgn \sigma) \sigma f$
Solution:
$Sf = \sum_{\sigma \in S_2} \sigma f = (2x+3y)+(2y+3x) = 5x + 5y$
Seems reasonable, as $\sigma f = f$ for any $\sigma$.
$Af = \sum_{\sigma \in S_2} (sgn \sigma) \sigma f = (1)(2x + 3y)+(-1)(2y+3x)=-x+y$
Also seems reasonable, as $\sigma f = (sgn \sigma)f$ for any $\sigma$.

Your calculation for $Sf$ is correct. However, it is not true that $\sigma f= f$ for any $\sigma$; i.e., $f$ is not symmetric. Can you see why?

2.
If $g(x,y)=x+y$, show $Ag=0$ and $Sg=g$.
Solution:
$Sg = \sum_{\sigma \in S_2} \sigma g = (x+y)+(y+x)=2x+2y$
Though I am supposed to show that $Sg=g$. Why does it? Of course $g$ is already symmetric, but doesn't the definition for $Sg$ give $2x+2y$? Which is also symmetric.
$Ag = \sum_{\sigma \in S_2} (sgn \sigma) \sigma g = (1)(x+y)+(-1)(y+x)=0$

Your calculations are correct here. There must have been a typo in the problem statement given to you by your professor. As you noted, $g$ is symmetric, so $\sigma g=g$ for all $\sigma\in S_{2}$. Hence $Sg = 2g.$

3.
If $f(x,y)=2x+3y$ and $g(x,y)=4x+5y$, find $f \otimes g$ and $f \wedge g$.
I am given, (for f k-linear and g l-linear)
$( f \otimes g )(v_1, ..., v_{k+l})= f(v_1,..., v_k)\cdot g(v_{k+1},..., v_{k+l})$
and
$f \wedge g = \frac{1}{k!l!}A(f \otimes g)$
Solution:
$( f \otimes g )(x, y, a, b) = (2x+3y)(4a+5b)=8xa+10xb+12ya+15yb$

$f \wedge g = \frac{1}{2!2!}A(8xa+10xb+12ya+15yb)$
I am unsure about both of these. If that's right, then next I'd find the alternating 4-linear function that comes from $8xa+10xb+12ya+15yb$.

This looks good so far. As you said, calculating $f \wedge g = \frac{1}{2!2!}A(8xa+10xb+12ya+15yb)$ still remains.

 

[joypav]

Your calculation for $Sf$ is correct. However, it is not true that $\sigma f= f$ for any $\sigma$; i.e., $f$ is not symmetric. Can you see why?

Of course, yes. Mistake on my part.. I meant my calculation for $Sf$ made sense as $\sigma Sf = Sf, \forall \sigma$.

This looks good so far. As you said, calculating $f \wedge g = \frac{1}{2!2!}A(8xa+10xb+12ya+15yb)$ still remains.

Okay, great.
Is this calculation as long as it seems? Aren't there 24 possible $\sigma$'s?