Gentlemen! I only returned to this today, but I get it now. Thanks! I'm not 
sure whom to mark as "best answer", though; both seem equally good.

john perry

On Saturday, January 24, 2015 at 2:54:22 PM UTC-6, Andrew Ohana wrote:
>
>
>
> On Fri, Jan 23, 2015 at 8:13 PM, john_perry_usm <[email protected] 
> <javascript:>> wrote:
>
>>
>> Try the following:
>>>
>>> sage: e = SymmetricFunctions(QQ).e() # construct the symmetric functions 
>>> with the e basis 
>>> sage: m = SymmetricFunctions(QQ).m() # ditto but with the monomial basis
>>> sage: m421 = m[4, 2, 1] # create the monomial you care about
>>> sage: e(m421) # coerce the monomial into the ring with the e basis
>>> e[3, 2, 1, 1] - 2*e[3, 2, 2] ...
>>>
>>
>> That's not obviously the same as the result I specified.
>>
>
> Well, it is the result in when working with (countably) infinitely many 
> variables. There is a standard map from these to symmetric polynomials in 3 
> variables (obtained by setting x_i = 0 for i > 3), however from what I can 
> tell, the only way sage implements this is with the expand method -- which 
> is not what you want.
>
> In the case of the elementary symmetric polynomials basis (and only that 
> basis), you can obtain the result that you are looking for by restricting 
> the size of the parts to the number of variables you are working with. In 
> sage this can be done with the following code (continuing from where I left 
> off before):
>
> sage: em421 = e(m421) # coerce the monomial into the e basis
> sage: em421.restrict_parts(3) # restrict the size of the parts to at most 3
> e[3, 2, 1, 1] - 2*e[3, 2, 2] - e[3, 3, 1]
>
>
> This agrees with the result that you got by hand since e[a, b, c, ...] = 
> e_a*e_b*e_c*...
>  
>
>> I left out that I know the number of variables I have available; in the 
>> example above, I have three variables, so I'm looking specifically for 
>> e1=x1+x2+x3, etc. I don't see how I get that from the expression e[3,2,1,1] 
>> - 2*e[3,2,2] ...
>>
>> I'm quite new to this topic, hence my lack of clarity. Sorry about this. 
>>
>
>> john perry
>>
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>
>
>
> -- 
> Andrew
>  

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