For square integrable functions $f,g$ of a real variable, the Cauchy-Schwarz inequality states that $$ \left(\int f(x)g(x)\,dx \right)^2 \le \int f(x)^2\,dx \int g(x)^2\,dx. $$

My question is: are there any known analogues of this for functions of Grassmann (or mixed) variables?

As a first check, let us see what happens for functions of a single Grassmann variable $\xi$. The functions $f$ and $g$ are then of the form $$ f(\xi) = f_0 + f_1\xi \qquad g(\xi) = g_0 + g_1\xi. $$ Computing the left hand side of our tentative "Fermionic Cauchy Schwarz" we have $$ \left(\int f(\xi)g(\xi) \,d\xi\right)^2 = \left(\int d\xi \, f_0g_0 + (g_0f_1+f_0g_1)\xi \,d\xi \right)^2 = g_0^2f_1^2 + 2f_0g_0f_1g_1 + f_0^2g_1^2 $$ whilst on the right hand side we have $$ \int f(\xi)^2\,d\xi \int g(\xi)^2 \, d\xi = \int f_0^2 + 2f_0f_1 \xi \, d\xi \int g_0^2 + 2g_0g_1 \xi \, d\xi = 4f_0g_0f_1g_1. $$ Putting the two sides together, taking $a = g_0f_1$ and $b = f_0g_1$ we have $$ \begin{split} &g_0^2f_1^2 + 2f_0g_0f_1g_1 + f_0^2g_1^2 \stackrel{?}{\le} 4f_0g_0f_1g_1\\ &g_0^2f_1^2 + f_0^2g_1^2 \stackrel{?}{\le} 2f_0g_0f_1g_1\\ &a^2 + b^2 \stackrel{?}{\le} 2ab\\ &a^2 + b^2 \stackrel{!}{\ge} 2ab \end{split} $$ So the "Fermionic Cauchy Schwarz" is just the elementary Young's inequality! Not too surprisingly, the Fermionic Cauchy Schwarz goes the other direction, which gels with the usual Grassmann experience as "normal mathematics, but upside-down and in a mirror".

My attempts at extending this to mixed functions have thus far been unsuccessful with the exception of functions of a supersymmetric variable (say $z = x^2+y^2+2\xi\eta$), which trivially gives the equality $f(0)^2g(0^2) = f(0)^2g(0)^2$.

In fact, even trying the above computation with just two Grassmann variables leads to an expression which is not an inequality in either direction. The hope is that perhaps this is fixable when working with the "correct" class of functions, say, analytic functions of a single variable applied to a Grassmann variable, or the product of a real function with an analytic function of a supersymmetric bosonic variable.

If anyone is aware of any references/paths forward/fundamental obstructions to the idea I would be very appreciative.


This seems like a futile exercise.

  1. First of all, we assume that $f$ and $g$ carry definite Grassmann parity. (This is a standard assumption in super-mathematics.)

  2. If we assume that $f$ is Grassmann-odd, then $f^2=0$, i.e. the construction is stillborn.

  3. So let's assume from now on that $f(\theta)=f_0+\theta f_1$ and $g(\theta)=g_0+\theta g_1$ are Grassmann-even functions of a Grassmann-odd indeterminate $\theta$. In other words, $f_0,g_0$ are Grassmann-even and $f_1,g_1$ are Grassmann-odd.

  4. Then the Berezin integral $I=\int\! d\theta~ f(\theta)g(\theta) $ is Grassmann-odd, and hence the square $I^2=0$, i.e. the construction is again stillborn.

  5. Moreover, $ \int\! d\theta ~f(\theta)^2 \int\! d\theta^{\prime} ~g(\theta^{\prime})^2$ is a product of two Grassmann-odd supernumbers, i.e. it is a soul-valued Grassmann-even supernumber, cf. e.g. my Phys.SE answer here. It is unclear how to introduce an ordering $\leq$ among soul-valued Grassmann-even supernumbers, and more generally, how to interpret soul-valued supernumbers.

  • $\begingroup$ Thanks for the help but I don't think this is correct! 1. Yes, if f is odd then certainly f^2 is 0. 2. I suppose we could assume this, but if f and g have both even and odd components, then both odd(f)odd(g) and even(f)even(g) terms will contribute to the integral on the LHS (assuming there is more than 1 Grassmann variable of course). 3. After integrating out all Grassmann components, we are always left with something which is real, so this certainly will not square to zero in general (see my example). 4. Again, int(f^2)int(g^2) will always be real, not grassmann even. $\endgroup$ – Andrew Swan Feb 22 '18 at 16:03
  • $\begingroup$ I update the answer. Note that the numbers have shifted. $\endgroup$ – Qmechanic Feb 22 '18 at 16:49
  • $\begingroup$ There is still the issue with 4 and 5. Berezin integration always produces a real number, not a Grassmann number, after all terms have been integrated out. $\endgroup$ – Andrew Swan Feb 22 '18 at 17:12
  • $\begingroup$ I think there is some confusion about the sorts of functions I am interested in: I am only ever considering real functions applied to Grassmann variables so that Berezin integration produces a real number and not a supernumber. $\endgroup$ – Andrew Swan Feb 22 '18 at 17:27
  • $\begingroup$ Aha. Then $f,g$ would generically not carry definite Grassmann parity. $\endgroup$ – Qmechanic Feb 22 '18 at 17:34

OK, it seems that OP additionally requires that the coefficients (of a superpolynomial) belong to the body $\mathbb{R}$ (rather than the soul).

  1. One may show that OP's opposite(!) Cauchy-Schwarz (CS) inequality $$\left(\int\!d^n\theta~ f(\theta)g(\theta)\right)^2 ~\geq~ \int\!d^n\theta~ f(\theta) \int\!d^n\theta^{\prime}~ g(\theta^{\prime})\tag{1}$$ holds if both $f(\theta)$ and $g(\theta)$ are a sum of $\leq 2$ monomials. E.g. ineq. (1) holds with $n=2$ and $$f(\theta)~=~f_0+\theta_1\theta_2 f_3, \qquad g(\theta)~=~g_0+\theta_1\theta_2 g_3, \qquad f_0,f_3,g_0,g_3~\in~ \mathbb{R}. \tag{2}$$ Ineq. (1) does not hold for more general polynomials.

  2. Another idea is to define a sesquilinear form $$\langle f, g\rangle~\propto~ \int\!d^n\theta^{\prime} ~\overline{f(\theta)} \wedge \star g(\theta),\tag{3}$$ with the help of the Hodge star operator $\star$ (including appropriate sign factors). This will automatically satisfy a usual CS inequality. (Here we assume that the Grassmann indeterminates $\theta_i$ are real.)


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